($\stackrel{\mathbf{´}}{\mathbf{\in }}\mathbf{,}\stackrel{\mathbf{´}}{\mathbf{\in }}\mathbf{\vee }{\stackrel{\mathbf{´}}{\mathit{q}}}_{\stackrel{\mathbf{ˇ}}{\mathit{k}}}$)-Uni-Intuitionistic Fuzzy Soft h-Ideals in Subtraction BG-Algebras

Abstract

The main purpose of the present paper is to introduced the notions of $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SAs$ in subtraction BG-algebras. We provide different characterizations and some equivalent conditions of $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SAs$ in terms of the level subsets of subtraction BG-algebras. It has been revealed that the $(\acute{q},\acute{q})$-$U{I}_{FS}SA$ are $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$ but the converse does not hold and an example is provided. We introduced $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}IDs$ and its some usual properties. In addition, $h^{-1}(\tilde{\mathfrak{N}}[\varsigma ])$ is $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$. Moreover, if $h^{-1}(\tilde{\mathfrak{N}}[\varsigma ])$ are an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$, then $\tilde{\mathfrak{N}}[\varsigma ]$ are an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$. Finally, we characterize $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$ which is a generalization of $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}{H}_{ID}$.

1. Introduction

Fuzzy sets can be considered as an expansion and ridiculous misrepresentation of the classical sets presented by Zadeh [1]. They tend to be best grasped with regard to set participation. Intuitionistic fuzzy sets have been presented by Atanassov [2]. In arithmetic, BCI and BCK algebras are mathematical designs in general variable-based math, which were presented by Imai et al. [3] in 1966; they portray fragments of propositional math including suggestions known as BCI and BCK algebras. Neggers et al. [4] presented and explored a class of algebras which is connected with a few classes of algebras of interest, such as BCI/BCI/BCK-algebras, and which appear to have rather pleasant properties without being unnecessarily convoluted in any case.

Kim et al. [5] presented another idea, called a BG-algebra, which is a speculation of B-algebra. Zarandi et al. [6] considered the intuitionistic fuzzification of the idea of subalgebras and ideals in BG-algebras and explored a portion of their properties. Senapati et al. [7] researched a few properties of the intuitionistic fuzzy ideals of BG-algebras. Ahn et al. [8] arranged the subalgebras by the group of the level subalgebras in BG-algebras. Another kind of fuzzy subgroup was presented by Bhakat et al. [9] utilizing the consolidated thoughts of “belongings” and the “quasi-coincidence” of fuzzy points. Summing up, the possibility of the quasi-coincident of a fuzzy point with a fuzzy subset was presented by Jun in [10]. Basnet et al. [11,12] defined the $(\in ,\in \vee q)$-fuzzy ideal of a BG/d-algebra and investigated quite a few of its useful properties.

Molodstov [13] developed the thought of soft sets in 1999 as an innovative mathematical mechanism for allocating with an ambiguity specifically devoid of the problems that are sure to plague the traditional theoretical approach. Maji et al. [14,15] worked on the theoretical study of intuitionistic fuzzy soft sets in detail. Balamurugan et al. (see, e.g., [16,17,18]) introduced anti-intuitionistic fuzzy soft ideals, $(\in ,\in \vee \widehat{q})$-bipolar fuzzy b-ideals of BG/BCK/BCI-algebras. Muhiuddin et al. (see, for e.g., [19,20]) introduced an m-polar fuzzy set theoretic approach to generalized ideals in BCK/BCI-algebras.

In this paper, we introduce the ideas of $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SAs$ of subtraction BG-algebra and investigate some of their usual properties. We introduce $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}IDs$ and its some usual properties. In addition, $h^{-1}(\tilde{\mathfrak{N}}[\varsigma ])$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$. Moreover, $h^{-1}(\tilde{\mathfrak{M}}[\varsigma ])$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$, then $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$. Finally, we characterize $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$, which is a generalization of $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}{H}_{ID}$.

This manuscript frequently uses the following symbols and their meanings, which we present in the following,

Table 1. List of symbols used in this article.

Symbol	Representation
$F_{S}S$	Fuzzy soft set
$U_{FS}SA$	Uni-fuzzy soft subalgebra
$I_{F}S$	Intuitionistic fuzzy set
$I_{FS}S$	Intuitionistic fuzzy soft set
$U{I}_{FS}SA$	Uni-intuitionistic fuzzy soft subalgebra
$I_{FS}SU$	Intuitionistic fuzzy soft subset
$I_{FS}P$	Intuitionistic fuzzy soft point
$U{I}_{FS}ID$	Uni-intuitionistic fuzzy soft ideal
$U{I}_{FS}{H}_{ID}$	Uni-intuitionistic fuzzy soft h-ideal

.

2. Preliminaries

This study focuses on the subtraction BG-algebras, which is a generalization of BG-algebras.

Definition 1.

A subtraction BG-algebra is defined to be a set $\mathfrak{Y}$ with a special element 0 and a binary operation “-” which fulfils the succeeding axioms:

$(a)\theta -\theta =0;$

$(b)\theta -0=\theta ;$

$(c)(\theta -\vartheta )-(0-\vartheta )=\theta$ for, all $\theta ,\vartheta \in \mathfrak{Y}$.For brevity, we, likewise, call $\mathfrak{Y}$ a subtraction BG-algebra.

Example 1.

Let $\mathfrak{Y}=\{0,\theta ,\vartheta ,\phi ,\eta \}$ with the following Cayley

Table 2. Example of $U{I}_{FS}SA$ subtraction BG-algebra.

−	0	$\theta$	$\vartheta$	$\phi$	$\varpi$
0	0	$\varpi$	$\phi$	$\vartheta$	$\theta$
$\theta$	$\theta$	0	$\varpi$	$\phi$	0
$\vartheta$	$\vartheta$	$\theta$	0	$\varpi$	$\phi$
$\phi$	$\phi$	$\vartheta$	$\theta$	0	$\varpi$
$\varpi$	$\varpi$	$\phi$	$\vartheta$	$\theta$	0

:

Then $\mathfrak{Y}$ is a subtraction BG-algebra.

Definition 2 

(see [8]). A non-empty subset $\mathfrak{B}$ is a subalgebra of $\mathfrak{Y}$ if it satisfies

$(i)0\in \mathfrak{B}$;

$(ii)\theta \in \mathfrak{B},\vartheta \in \mathfrak{B}\Rightarrow \theta -\vartheta \in \mathfrak{B}$, for any $\theta ,\vartheta \in \mathfrak{Y}$.

Definition 3 

(see [13]). A pair $(\mathfrak{N},\mathfrak{B})$ is a soft set over $\mathfrak{Y}$ if and only if $\mathfrak{N}:\mathfrak{B}\to \mathfrak{P}(\mathfrak{Y})$.

Definition 4 

(see [1]). A fuzzy set $\mathfrak{B}$ is an object of the system $\mathfrak{B}=(\theta ,{\varkappa }_{\mathfrak{B}}(\theta )):\theta \in \mathfrak{Y}$, where ${\varkappa }_{\mathfrak{B}}:\mathfrak{Y}\to [0,1]$ with $0\le {\varkappa }_{\mathfrak{B}}\le 1$, for all $\theta \in \mathfrak{Y}$. The number ${\varkappa }_{\mathfrak{B}}$ denotes the degree of members of the element $\theta \in \mathfrak{B}$.

Definition 5. 

(see [14]). Let $\mathfrak{Y}$ be a set, and let $\mathfrak{E}$ be a set of factors. If $\mathfrak{F}(\mathfrak{Y})$ is the set of all fuzzy sets, then $(\tilde{\mathfrak{N}},\mathfrak{B})$ is $F_{S}S$ over $\mathfrak{Y}$, and $\mathfrak{B}\subseteq \mathfrak{E}$, where $\tilde{\mathfrak{N}}:\mathfrak{B}\to \mathfrak{F}(\mathfrak{Y})$.

Definition 6 

(see [16]). Let $(\tilde{\mathfrak{N}},\mathfrak{B})$ be $F_{S}S$. We say that $(\tilde{\mathfrak{N}},\mathfrak{B})$ is an $U_{FS}SA$ if $\tilde{\mathfrak{N}}[\varsigma ]$ is an $F_{S}SU$ which satisfies

(a) ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(0)\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )$;

(b) ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$, for all $\theta \in \mathfrak{N},\varsigma \in \mathfrak{B}$.

Definition 7 

(see [2]). An $I_{F}S$$\mathfrak{B}$ of $\mathfrak{Y}$ is of the form $\mathfrak{B}=\{(\theta ,{\varkappa }_{\mathfrak{B}}(\theta ),{\varpi }_{\mathfrak{B}}(\theta )):\theta \in \mathfrak{Y}\}$, where ${\varkappa }_{\mathfrak{B}},{\varpi }_{\mathfrak{B}}:\mathfrak{Y}\to [0,1]$ with $0\le {\varkappa }_{\mathfrak{B}}(\theta )+{\varpi }_{\mathfrak{B}}(\theta )\le 1$, for all $\theta \in \mathfrak{Y}$.

Definition 8 

(see [15]). Let $\mathfrak{Y}$ be a set and let $\mathfrak{E}$ be a set of factors. If $\mathfrak{IF}(\mathfrak{Y})$ is the set of all $I_{F}Ss$ then $(\tilde{\mathfrak{N}},\mathfrak{B})$ is an $I_{FS}S$, where $\tilde{\mathfrak{N}}:\mathfrak{B}\to \mathfrak{IF}(\mathfrak{Y})$.

Definition 9 

(see [16]). Let $(\tilde{\mathfrak{N}},\mathfrak{B})$ be an $I_{FS}S$. Then we say that $(\tilde{\mathfrak{N}},\mathfrak{B})$ is an $U{I}_{FS}SA$ if $\tilde{\mathfrak{N}}[\varsigma ]$ is an $I{F}_{S}SU$ which satisfies

(a) ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(0)\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(0)\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta );$

(b) ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$;

(c) ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$, for all $\theta \in \mathfrak{N},\varsigma \in \mathfrak{B}$.

Example 2.

Consider the subtraction BG-algebra $\mathfrak{Y}=\{0,\theta ,\vartheta \}$ which has the Cayley

Table 3. Cayley table for $U{I}_{FS}SA$.

-	0	$\theta$	$\vartheta$
0	0	$\theta$	$\vartheta$
$\theta$	$\theta$	0	$\theta$
$\vartheta$	$\vartheta$	$\vartheta$	0

given below:

Define an $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ given by

$${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(0)={\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )=0.22,{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )=0.42,$$

and

$${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(0)={\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )=0.72,{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )=0.32.$$

Then $\tilde{\mathfrak{N}}[\varsigma ]$ is an $U{I}_{F}SA$. Hence, $(\tilde{\mathfrak{N}},\mathfrak{B})$ is an $U{I}_{FS}SA$.

3. $(\stackrel{\mathbf{´}}{\mathbf{\in }}\mathbf{,}\stackrel{\mathbf{´}}{\mathbf{\in }}\mathbf{\vee }\stackrel{\mathbf{´}}{\mathit{q}})$-Uni-Intuitionistic Fuzzy Soft Subalgebras

Definition 10.

An $I_{FS}P$${\theta }_{\tilde{\mathfrak{t}}}$ is said to belong to (respectively, be quasi-coincident with) an $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ written as ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }\tilde{\mathfrak{N}}[\varsigma ]$ if ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\tilde{\mathfrak{t}}>1$ or ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge \tilde{\mathfrak{t}}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\tilde{\mathfrak{t}}<1$ or ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\le \tilde{\mathfrak{t}}$.

Definition 11.

An $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$ of $\mathfrak{Y}$ if ${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ implies ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}$ implies ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }{\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}$, for all $\theta ,\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$.

Definition 12.

An $I_{FS}P$${\theta }_{\tilde{\mathfrak{t}}}$ is said to belong to (respectively, be quasi-coincident with) an $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ written as ${\theta }_{\tilde{\mathfrak{t}}}\acute{q}\tilde{\mathfrak{N}}[\varsigma ]$ (respectively, ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }\tilde{\mathfrak{N}}[\varsigma ]$) if ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\tilde{\mathfrak{t}}>1$ or ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge \tilde{\mathfrak{t}}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\tilde{\mathfrak{t}}<1$ or ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\le \tilde{\mathfrak{t}}$. If ${\theta }_{\tilde{\mathfrak{t}}}\acute{q}\tilde{\mathfrak{N}}[\varsigma ]$ or ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }\tilde{\mathfrak{N}}[\varsigma ]$, then ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }\vee \acute{q}\tilde{\mathfrak{N}}[\varsigma ]$.

Definition 13.

An $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$ of $\mathfrak{Y}$ if ${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ implies ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}$ implies ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}$, for all $\theta ,\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$.

Definition 14.

An $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$ if it satisfies the following conditions:

$(i){\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )$;

$(ii){\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )$.

Theorem 1.

$\tilde{\mathfrak{N}}[\varsigma ]$ is an $U{I}_{FS}SA$ iff $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$.

Proof. 

Let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $U{I}_{FS}SA$. Then

${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$.

Let $\theta ,\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$ s.t ${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )$, wherever $\tilde{\mathfrak{t}},\tilde{\mathfrak{s}}\in (0,1)$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\le \tilde{\mathfrak{t}},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{s}}$, and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge \tilde{\mathfrak{t}},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge \tilde{\mathfrak{s}}$. Now ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}\Rightarrow {(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}\Rightarrow {(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Therefore, $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$.

Conversely, let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$. To verify that $\tilde{\mathfrak{N}}[\varsigma ]$ is an $U{I}_{FS}SA$, let $\theta ,\vartheta \in \mathfrak{Y}$, $\tilde{\mathfrak{t}}={\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )$ and $\tilde{\mathfrak{s}}={\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\le \tilde{\mathfrak{t}},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{s}}$ implies ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Therefore, ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$, i.e., ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$. Therefore,

$${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta ).$$

(1)

Again, let $\theta ,\vartheta \in \mathfrak{Y}$ and $\tilde{\mathfrak{t}}={\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )$, $\tilde{\mathfrak{s}}={\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge \tilde{\mathfrak{t}}$, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge \tilde{\mathfrak{s}}$ implies ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Thus, ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$, i.e., ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$. Therefore,

$${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta ).$$

(2)

Hence, (1) and (2) $\Rightarrow \tilde{\mathfrak{N}}[\varsigma ]$ is an $U{I}_{FS}SA$. □

Definition 15.

An $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{q},\acute{q})$-$U{I}_{FS}SA$ of $\mathfrak{Y}$ if ${\theta }_{1-\tilde{\mathfrak{t}}-\breve{\epsilon }},{\vartheta }_{1-\tilde{\mathfrak{s}}-\breve{\epsilon }}\acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ implies ${(\theta -\vartheta )}_{(1-\tilde{\mathfrak{t}}-\breve{\epsilon })\vee (1-\tilde{\mathfrak{s}}-\breve{\epsilon })}\acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\theta }_{1-\tilde{\mathfrak{t}}+\breve{\epsilon }},{\vartheta }_{1-\tilde{\mathfrak{s}}+\breve{\epsilon }}\acute{q}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ implies ${(\theta -\vartheta )}_{(1-\tilde{\mathfrak{t}}+\breve{\epsilon })\wedge (1-\tilde{\mathfrak{s}}+\breve{\epsilon })}\acute{q}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$, for all $\theta ,\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$.

Theorem 2.

Every $(\acute{q},\acute{q})$-$U{I}_{FS}SA$ is an $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$.

Proof. 

Let $\tilde{\mathfrak{N}}[\varsigma ]$ be a $(\acute{q},\acute{q})$-$U{I}_{FS}SA$. Let $\theta ,\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$ such that ${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\le t,{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le s$ implies ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )-\breve{\epsilon }<\tilde{\mathfrak{t}}$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )-\breve{\epsilon }<\tilde{\mathfrak{s}}$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+1-\tilde{\mathfrak{t}}-\breve{\epsilon }<1$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )+1-\tilde{\mathfrak{s}}-\breve{\epsilon }<1$, i.e., ${\theta }_{1-\tilde{\mathfrak{t}}-\breve{\epsilon }}\acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\vartheta }_{1-\tilde{\mathfrak{s}}-\breve{\epsilon }}\acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$.

Since $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{q},\acute{q})$-$U{I}_{FS}SA$, we have ${(\theta -\vartheta )}_{(1-\breve{\epsilon }-\tilde{\mathfrak{t}})\vee (1-\breve{\epsilon }-\tilde{\mathfrak{s}})}\acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$

$\Rightarrow {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+(1-\tilde{\mathfrak{t}}-\breve{\epsilon })\vee (1-\tilde{\mathfrak{s}}-\breve{\epsilon })<1$

$\Rightarrow {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+1-(\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}})-\breve{\epsilon }<1$

$\Rightarrow {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<(\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}})+\breve{\epsilon }$

$\Rightarrow {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$

$\Rightarrow {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}<\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$

$\Rightarrow {(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$.

Therefore,

$${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow {(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(3)

Again, let $\theta ,\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$ such that ${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge \tilde{\mathfrak{t}},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge \tilde{\mathfrak{s}}$ implies ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\breve{\epsilon }>\tilde{\mathfrak{t}}$, and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )+\breve{\epsilon }>\tilde{\mathfrak{s}}$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\breve{\epsilon }-\tilde{\mathfrak{t}}+1>1$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )+\breve{\epsilon }-\tilde{\mathfrak{s}}+1>1$, i.e., ${\theta }_{\breve{\epsilon }-\tilde{\mathfrak{t}}+1}\acute{q}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\vartheta }_{\breve{\epsilon }-\tilde{\mathfrak{s}}+1}\acute{q}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$.

Since $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$, we have

${(\theta -\vartheta )}_{(\breve{\epsilon }-\tilde{\mathfrak{t}}+1)\wedge (\breve{\epsilon }-\tilde{\mathfrak{s}}+1)}\acute{q}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$

$\Rightarrow {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+(\breve{\epsilon }-\tilde{\mathfrak{t}}+1)\wedge (\breve{\epsilon }-\tilde{\mathfrak{s}}+1)<1$

$\Rightarrow {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\breve{\epsilon }+1-(\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}})>1$

$\Rightarrow {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )>(\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}})-\breve{\epsilon }$

$\Rightarrow {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )>\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$

$\Rightarrow {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )>\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}>\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$

$\Rightarrow {(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$.

Therefore,

$${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow {(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(4)

Hence, (3) and (4) $\Rightarrow \tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$. □

The following example shows the converse of Theorem 2 may not be true.

Example 3.

Consider a subtraction BG-algebra $\mathfrak{Y}=\{0,\theta ,\vartheta ,\phi \}$ with the following Cayley

Table 4. Illustration of converse of Theorem 2.

−	0	$\theta$	$\vartheta$	$\phi$
0	0	$\theta$	$\vartheta$	$\phi$
$\theta$	$\theta$	0	$\phi$	$\vartheta$
$\vartheta$	$\vartheta$	$\phi$	0	$\theta$
$\phi$	$\phi$	$\vartheta$	$\theta$	0

:

Define $\tilde{\mathfrak{N}}[\varsigma ]$ by

−	0	$\theta$	$\vartheta$	$\phi$
${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$	0.52	0.52	0.56	0.56
${\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}$	0.41	0.41	0.34	0.34

Then $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$, but it is not a $(\acute{q},\acute{q})$-$U{I}_{FS}SA$. As, if $\tilde{\mathfrak{t}}=0.71$, $\tilde{\mathfrak{s}}=0.61$, then ${\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\tilde{\mathfrak{t}}={\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}(\phi )+0.71=0.31+0.71=1.02>1$, and ${\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )+\tilde{\mathfrak{s}}={\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+0.61=0.41+0.61=1.02>1$, but ${\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}=0.34+0.71\wedge 0.61=0.34+0.61=0.95<1.$

Theorem 3.

An $I_{F}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$ if and only if

(i) ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2$;

(ii) ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2$;

for all $\theta ,\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$.

Proof. 

(i) First, let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$.

Case 1: Let ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )>1/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2={\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. If possible, let ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. Choose $\tilde{\mathfrak{t}}$ as a real number such that ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )>\tilde{\mathfrak{t}}>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )<\tilde{\mathfrak{t}},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )<\tilde{\mathfrak{t}}$. Therefore, ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},{\vartheta }_{\tilde{\mathfrak{t}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. However, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )>\tilde{\mathfrak{t}}$ implies ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}}\acute{\notin }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}>2\tilde{\mathfrak{t}}$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}>2({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta ))>2\times 1/2=1$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}>1$, which fails due to the fact that $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$. Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2$.

Case 2: Let ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le 1/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )=1/2$. If possible, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2=1/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\le 1/2$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le 1/2$. Therefore, ${\theta }_{1/2},{\vartheta }_{1/2}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. However, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )>1/2$; therefore, ${\theta }_{1/2}\notin {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+1/2>1/2+1/2=1$, which again fails due to the fact that $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$. Hence, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le 1/2={\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2$.

Converse Part:

Let

$${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2.$$

(5)

Let $\theta ,\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$ such that ${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\le \tilde{\mathfrak{t}}$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{s}}$. Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$. By (5), ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le 1/2\vee \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$, let $1/2\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$. Then $1/2\vee \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}=\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$ implies ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$. Therefore,

$${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(6)

Again, let $1/2<\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$. Then $1/2\vee \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}=1/2$ implies ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}\vee 1/2=1/2$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}<1/2+1/2=1$. Therefore,

$${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(7)

From Equations (6) and (7), we have

$${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow {(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(8)

Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$.

$(ii)$ First, let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$.

Case 1: Let ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )<1/2$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2={\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. If possible, let ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. Then let $\tilde{\mathfrak{t}}$ be a real number such that ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<\tilde{\mathfrak{t}}<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )>\tilde{\mathfrak{t}},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )>\tilde{\mathfrak{t}}$. Therefore, ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]},{\vartheta }_t\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. However, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<\tilde{\mathfrak{t}}$. Then ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}}\acute{\notin }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\tilde{\mathfrak{t}}=2\tilde{\mathfrak{t}}$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}<2({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta ))<2\times 1/2=1$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}<1$, which fails due to the fact that $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$. Therefore, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )={\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2$.

Case 2: Let ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )=1/2$. If possible, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2=1/2$, then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge 1/2$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge 1/2$. Therefore, ${\theta }_{1/2},{\vartheta }_{1/2}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. However, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<1/2$. Then ${(\theta -\vartheta )}_{1/2}\acute{\notin }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+1/2<1/2+1/2=1$, which again fails due to the fact that $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$. Hence, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge 1/2={\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2$.

Converse Part:

Let

$${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2.$$

(9)

Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge t$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge \tilde{\mathfrak{s}}$. Therefore, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$. By (9), ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge 1/2\vee \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$. Let $1/2\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$. Then $1/2\wedge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}=\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$. Therefore,

$${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(10)

Again, let $1/2<\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$. Then $1/2\wedge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}=1/2$. Therefore, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}\wedge 1/2=0$, i.e., ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}>1/2+1/2=1$. Therefore,

$${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{q}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(11)

From (10) and (11), we have

$${\theta }_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow {(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(12)

Therefore, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$. Hence, (8) and (12) $\Rightarrow \tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$. □

Theorem 4.

An $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$, and if ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )>1/2,{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )<1/2$, then $\tilde{\mathfrak{N}}[\varsigma ]$ is also an $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$.

Proof. 

Let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$ of $\mathfrak{N}$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )>1/2$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )<1/2$. Let ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},{\vartheta }_s\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then $1/2<{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\le \tilde{\mathfrak{t}}$ and $1/2<{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{s}}$. Therefore, $\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}>1/2$. In addition, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )>1/2$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}>1/2+1/2=1$. Since ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$, we have either ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}<1or{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}\Rightarrow {(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Therefore,

$${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow {(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(13)

Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$.

Again, let ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then $\tilde{\mathfrak{t}}\le {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )<1/2$ and $\tilde{\mathfrak{s}}\le {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )<1/2$. Therefore, $\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}<1/2$ and also ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<1/2$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}<1/2+1/2=1$. Since ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$, we have either ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}>1$ or ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}\Rightarrow {(\theta -\vartheta )}_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Therefore,

$${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow {(\theta -\vartheta )}_{t\wedge s}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(14)

Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$. Hence, (13) and (14) $\Rightarrow \tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$. □

Definition 16.

Let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $I_{FS}SU$ of $\mathfrak{Y}$ and $\tilde{\mathfrak{t}}\in (0,1]$. Then let ${({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}=\{\theta :{\theta }_{\tilde{\mathfrak{t}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\}=\{\theta :{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge \tilde{\mathfrak{t}}\}$, ${\langle {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\rangle }_{\tilde{\mathfrak{t}}}=\{\theta :{\theta }_{\tilde{\mathfrak{t}}}\acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\}=\{\theta :{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\tilde{\mathfrak{t}}>1\}$, ${[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}=\{\theta :{\theta }_{\tilde{\mathfrak{t}}}\acute{\in }\vee \acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\}=\{\theta :{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\ge \tilde{\mathfrak{t}}\}$, where ${({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}$, ${\langle {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\rangle }_t$, and ${[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$ is $\tilde{\mathfrak{t}}$-level set, $\acute{q}$-level set, and $(\acute{\in }\vee \acute{q})$-level set of ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Clearly, ${[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}={\langle {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\rangle }_{\tilde{\mathfrak{t}}}\cup {({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}$, and ${({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}=\{\theta :{\theta }_{\tilde{\mathfrak{t}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\}=\{\theta :{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\le \tilde{\mathfrak{t}}\}$, ${\langle {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\rangle }_{\tilde{\mathfrak{t}}}=\{\theta :{\theta }_{\tilde{\mathfrak{t}}}q{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\}=\{\theta :{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}+\tilde{\mathfrak{t}}<1\}$, ${[{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}=\{\theta :{\theta }_{\tilde{\mathfrak{t}}}\acute{\in }\vee \acute{q}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\}=\{\theta :{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\le \tilde{\mathfrak{t}}or{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\tilde{\mathfrak{t}}<1\}$, where ${({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}$, ${\langle {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\rangle }_{\tilde{\mathfrak{t}}}$, and ${[{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$ is $(\acute{\in }\vee \acute{q})$ are the $\tilde{\mathfrak{t}}$-level set, $\acute{q}$-level set, and $(\acute{\in }\vee \acute{q})$-level set of ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Clearly, ${[{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}={\langle {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\rangle }_{\tilde{\mathfrak{t}}}\cup {({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}.$

Theorem 5.

An $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$ if and only if the sets ${({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}=\{\theta :{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )<\tilde{\mathfrak{t}},\tilde{\mathfrak{t}}\in (1/2,1],{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(0)<\tilde{\mathfrak{t}}\}$ and ${({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{s}}}=\{\theta :{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge \tilde{\mathfrak{s}},\tilde{\mathfrak{s}}\in (0,1/2),{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(0)\ge \tilde{\mathfrak{s}}\}$ are subalgebras of $\mathfrak{Y}$.

Proof. 

Suppose $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$. Then $0\acute{\in }{({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}},0\acute{\in }{({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{s}}}$. Let $\theta ,\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$ s.t $\theta ,\vartheta \acute{\in }{({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}$ where $\tilde{\mathfrak{t}}\in (1/2,1]$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )<\tilde{\mathfrak{t}},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )<\tilde{\mathfrak{t}}$. Using Theorem, $3.4$, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2<\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}\vee 1/2=\tilde{\mathfrak{t}}$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<\tilde{\mathfrak{t}},$ i.e., $\theta -\vartheta \acute{\in }{({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}$. Therefore, $\theta ,\vartheta \acute{\in }{({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}\Rightarrow \theta -\vartheta \acute{\in }{({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}$. Hence, ${({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}$ is a subalgebra of $\mathfrak{Y}$.

Next, let $\theta ,\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$ such that $\theta ,\vartheta \acute{\in }{({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{s}}}$ where $\tilde{\mathfrak{s}}\in (0,1/2)$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge \tilde{\mathfrak{s}},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge \tilde{\mathfrak{s}}$. Using Theorem $3.4$, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2>\tilde{\mathfrak{s}}\wedge \tilde{\mathfrak{s}}\wedge 1/2=\tilde{\mathfrak{s}}$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge \tilde{\mathfrak{s}}$, i.e., $\theta -\vartheta \acute{\in }{({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{s}}}$. Therefore, $\theta ,\vartheta \acute{\in }{({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{s}}}\Rightarrow \theta -\vartheta \acute{\in }{({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{s}}}$. Hence, ${({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{s}}}$ is a subalgebra of $\mathfrak{Y}$.

Conversely, $\tilde{\mathfrak{N}}[\varsigma ]$ is an $I_{FS}SU$ and ${({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}=\{\theta :{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )<\tilde{\mathfrak{t}},$ where $\tilde{\mathfrak{t}}\in (1/2,1]\}$ and ${({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{s}}}=\{\theta :{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge \tilde{\mathfrak{s}},$ where $\tilde{\mathfrak{s}}\in (0,1/2)\}$ are subalgebras of $\mathfrak{Y}$. To prove $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$, suppose $\tilde{\mathfrak{N}}[\varsigma ]$ is not an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$, there exists $\theta ,\vartheta \in \mathfrak{Y}$ s.t at least one of ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2$ hold. Suppose ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2$ holds. Let $\tilde{\mathfrak{t}}=\{{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2)\}/2$. Then $\tilde{\mathfrak{t}}\in (1/2,1]$ and

$${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )>\tilde{\mathfrak{t}}>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2.$$

(15)

Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )<\tilde{\mathfrak{t}},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )<\tilde{\mathfrak{t}}$. So, $\theta \acute{\in }{({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}},\vartheta \acute{\in }{({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}$, i.e., $\theta -\vartheta \acute{\in }{({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}$.

Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<\tilde{\mathfrak{t}}$, which contradicts (15). Hence,

$${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee 1/2.$$

(16)

Next, let ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2$ hold. Let $\tilde{\mathfrak{s}}=\{{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )+({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2)\}/2.$ Then $\tilde{\mathfrak{s}}\in (0,1/2)$ and

$${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )<\tilde{\mathfrak{s}}<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2.$$

(17)

Therefore, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )>\tilde{\mathfrak{s}},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )>\tilde{\mathfrak{s}}$. Thus, $\theta \acute{\in }{({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{s}}},\vartheta \acute{\in }{({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{s}}}$, i.e., $\theta -\vartheta \acute{\in }{({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{s}}}$. Therefore, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )>\tilde{\mathfrak{s}}$, which contradicts (17). Hence,

$${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge 1/2.$$

(18)

Hence, (16) and (18) $\Rightarrow \tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SA$. □

4. Homomorphism of $(\stackrel{\mathbf{´}}{\mathbf{\in }}\mathbf{,}\stackrel{\mathbf{´}}{\mathbf{\in }}\mathbf{\vee }\stackrel{\mathbf{´}}{{q}})$-Uni-Intuitionistic Fuzzy Soft Ideals

Definition 17.

Let $\mathfrak{Y}$ and ${\mathfrak{Y}}^{\prime }$ be two subtraction BG-algebras. Then $h:\mathfrak{Y}\to {\mathfrak{Y}}^{\prime }$ is said to be a homomorphism if $h(\theta -\vartheta )=h(\theta )-h(\vartheta )$, for all $\theta ,\vartheta \in \mathfrak{Y}.$

Definition 18.

A $I{F}_{S}SU$${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ of $\mathfrak{Y}$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$ of $\mathfrak{Y}$ if ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\vee s}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\wedge \tilde{s}}\acute{\in }{\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}$, for all $\theta ,\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$.

Definition 19.

An $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$ if it fulfils the succeeding conditions:

(i) ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta ),{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\le \tilde{\mathfrak{t}},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le s\Rightarrow {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$.

(ii) ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta ),{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\vartheta )\ge \tilde{\mathfrak{t}},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge \tilde{\mathfrak{s}}\Rightarrow {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$.

Theorem 6.

Let $h:\mathfrak{Y}\to {\mathfrak{Y}}^{\prime }$ be a homomorphism and $\mathfrak{Y},{\mathfrak{Y}}^{\prime }$ be two subtraction BG-algebras. If $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$, then $h^{-1}(\tilde{\mathfrak{N}}[\varsigma ])$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$.

Proof. 

Let $h^{-1}(\tilde{\mathfrak{N}}[\varsigma ])(\theta )=h^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})(\theta )$ be defined as $h^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})(\theta )=$$({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})(h^{-1}(\theta ))$. Let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$ of ${\mathfrak{Y}}^{\prime }$ and let $\theta ,\vartheta \in \mathfrak{Y}$ s.t ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{h}^{-1}(\tilde{\mathfrak{N}}[\varsigma ])=h^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})=(h^{-1}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},h^{-1}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})$. Then ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{h}^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})$ and ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{h}^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})$.

Case 1: Let ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{h}^{-1}({\sigma }_{\tilde{\mathfrak{N}}[\varsigma ]})$

$\Rightarrow h^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})(\theta -\vartheta )\le \tilde{\mathfrak{t}}$ and $h^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})(\vartheta )\le \tilde{\mathfrak{s}}$

$\Rightarrow {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}h(\theta -\vartheta )\le \tilde{\mathfrak{t}}$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}h(\vartheta )\le \tilde{\mathfrak{s}}$

$\Rightarrow {(h(\theta -\vartheta ))}_{\tilde{\mathfrak{t}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma \tilde{\mathfrak{s}}]}$ and ${(h(\vartheta ))}_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$

$\Rightarrow {(h(\theta )-h(\vartheta ))}_{\tilde{\mathfrak{t}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${(h(\vartheta ))}_{\tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$

$\Rightarrow {(h(\theta ))}_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$

$\Rightarrow {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\theta ))\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$ or ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\theta ))+\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}<1$

$\Rightarrow h^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta ))\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$ or $h^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta ))+\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}<1$

$\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }{h}^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})$ or ${\theta }_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }\acute{q}{h}^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})$

$\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{h}^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})$.

Therefore,

$${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{h}^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{h}^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}).$$

(19)

Case 2: Let ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{h}^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})$

$\Rightarrow h^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})(\theta -\vartheta )\ge \tilde{\mathfrak{t}}$ and $h^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})(\vartheta )\ge \tilde{\mathfrak{s}}$

$\Rightarrow {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}h(\theta -\vartheta )\ge \tilde{\mathfrak{t}}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}h(\vartheta )\ge \tilde{\mathfrak{s}}$

$\Rightarrow {(h(\theta -\vartheta ))}_{\tilde{\mathfrak{t}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${(h(\vartheta ))}_{\tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$

$\Rightarrow {(h(\theta )-h(\vartheta ))}_{\tilde{\mathfrak{t}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${(h(\vartheta ))}_{\tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$

$\Rightarrow {(h(\theta ))}_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$

$\Rightarrow {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\theta ))\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$ or ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\theta ))+\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}>1$

$\Rightarrow h^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta ))\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$ or $h^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta ))+\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}>1$

$\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }{h}^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})$ or ${\theta }_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }\acute{q}{h}^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})$

$\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{h}^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})$.

Therefore,

$${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{h}^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{h}^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}).$$

(20)

Hence (19) and (20) $\Rightarrow h^{-1}(\tilde{\mathfrak{N}}[\varsigma ])$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$. □

Theorem 7.

Let $h:\mathfrak{Y}\to {\mathfrak{Y}}^{\prime }$ be an onto homomorphism and $\mathfrak{Y},{\mathfrak{Y}}^{\prime }$ be two subtraction BG-algebras. If $\tilde{\mathfrak{N}}[\varsigma ]$ is an $I_{FS}SU$ s.t $h^{-1}(\tilde{\mathfrak{N}}[\varsigma ])$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$, then $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$.

Proof. 

Let ${\theta }^{\prime },{\vartheta }^{\prime }\in {\mathfrak{Y}}^{\prime }$ and $\varsigma \in \mathfrak{B}$ s.t ${({\theta }^{\prime }-{\vartheta }^{\prime })}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}^{\prime }\acute{\in }\tilde{\mathfrak{N}}[\varsigma ]$. ${({\theta }^{\prime }-{\vartheta }^{\prime })}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}^{\prime }\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${({\theta }^{\prime }-{\vartheta }^{\prime })}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}^{\prime }\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}({\theta }^{\prime }-{\vartheta }^{\prime })\le \tilde{\mathfrak{t}}$, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}({\vartheta }^{\prime })\le \tilde{\mathfrak{s}}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}({\theta }^{\prime }-{\vartheta }^{\prime })\ge \tilde{\mathfrak{t}},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}({\vartheta }^{\prime })\ge \tilde{\mathfrak{s}}$. Since h is onto, there exists $\theta ,\vartheta \in \mathfrak{Y}$ s.t $h(\theta )={\theta }^{\prime },h(\vartheta )={\vartheta }^{\prime }$, $h(\theta -\vartheta )=h(\theta )-h(\vartheta )={\theta }^{\prime }-{\vartheta }^{\prime }$.

Let ${({\theta }^{\prime }-{\vartheta }^{\prime })}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}^{\prime }\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\theta -\vartheta ))\le \tilde{\mathfrak{t}}$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\vartheta ))\le \tilde{\mathfrak{s}}$ imply $h^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})(\theta -\vartheta )\le \tilde{\mathfrak{t}}$ and $h^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})(\vartheta )\le \tilde{\mathfrak{s}}$. Thus, ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}}\acute{\in }{h}^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})$ and ${\vartheta }_{\tilde{\mathfrak{s}}}\acute{\in }{h}^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})$, i.e., ${\theta }_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{h}^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})$. It follows that $h^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})(\theta )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$ or $h^{-1}({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})(\theta )+\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}<1$ imply ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\theta ))\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$ or ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\theta ))+\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}<1$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}({\theta }^{\prime })\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$ or ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}({\theta }^{\prime })+\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}<1$, i.e., ${\theta }_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}^{\prime }\acute{\in }\vee \acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Therefore,

$${({\theta }^{\prime }-{\vartheta }^{\prime })}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}^{\prime }\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}}^{\prime }\acute{\in }\vee \acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(21)

Let ${({\theta }^{\prime }-{\vartheta }^{\prime })}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}^{\prime }\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\theta -\vartheta ))\ge \tilde{\mathfrak{t}}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\vartheta ))\ge \tilde{\mathfrak{s}}$ imply $h^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})(\theta -\vartheta )\ge \tilde{\mathfrak{t}}$ and $h^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})(\vartheta )\ge \tilde{\mathfrak{s}}$. Thus, ${(\theta -\vartheta )}_{\tilde{\mathfrak{t}}}\acute{\in }{h}^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})$ and ${\vartheta }_{\tilde{\mathfrak{t}}}\acute{\in }{h}^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})$, i.e., ${\theta }_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}\acute{\in }\vee \acute{q}{h}^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})$. It follows that $h^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})(\theta )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$ or $h^{-1}({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})(\theta )+\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}>1$ imply ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\theta ))\ge \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$ or ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(h(\theta ))+\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}>1$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}({\theta }^{\prime })\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$ or ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}({\theta }^{\prime })+\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}>1$, i.e., ${\theta }_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}^{\prime }\acute{\in }\vee \acute{q}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Therefore,

$${({\theta }^{\prime }-{\vartheta }^{\prime })}_{\tilde{\mathfrak{t}}},{\vartheta }_{\tilde{\mathfrak{s}}}^{\prime }\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow {\theta }_{\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}}^{\prime }\acute{\in }\vee \acute{q}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(22)

Hence, (21) and (22) $\Rightarrow \tilde{\mathfrak{N}}[\varsigma ]$ is also an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$. □

5. $(\stackrel{\mathbf{´}}{\mathbf{\in }}\mathbf{,}\stackrel{\mathbf{´}}{\mathbf{\in }}\mathbf{\vee }{\stackrel{\mathbf{´}}{\mathit{q}}}_{\stackrel{\mathbf{ˇ}}{\mathit{k}}})$-Uni-Intuitionistic Fuzzy Soft h-Ideals

Definition 20.

An $I_{FS}P$${\theta }_{\tilde{\mathfrak{t}}}$ is said to belong to (respectively, be quasi-coincident with) an $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ written as ${\theta }_{\tilde{\mathfrak{t}}}{\acute{q}}_{\check{k}}\tilde{\mathfrak{N}}[\varsigma ]$ (respectively, ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }\tilde{\mathfrak{N}}[\varsigma ]$) if ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\tilde{\mathfrak{t}}+\check{k}>1$ or ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\ge \tilde{\mathfrak{t}}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )+\tilde{\mathfrak{t}}+\check{k}<1$ or ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta )\le \tilde{\mathfrak{t}}$. If ${\theta }_{\tilde{\mathfrak{t}}}{\acute{q}}_{\check{k}}\tilde{\mathfrak{N}}[\varsigma ]$ or ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }\tilde{\mathfrak{N}}[\varsigma ]$, then ${\theta }_{\tilde{\mathfrak{t}}}\acute{\in }\vee {\acute{q}}_{\check{k}}\tilde{\mathfrak{N}}[\varsigma ]$.

Definition 21.

An $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in })$-$U{I}_{FS}{H}_{ID}$ if it satisfies the following conditions:

(i) $((\theta -(\vartheta -\phi )),\tilde{\mathfrak{t}}),(\vartheta ,\tilde{\mathfrak{s}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow (\theta -\phi ,\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$;

(ii) $((\theta -(\vartheta -\phi )),\tilde{\mathfrak{t}}),(\vartheta ,\tilde{\mathfrak{s}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow (\theta -\phi ,\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$,

for every $\theta ,\vartheta ,\phi \in \mathfrak{Y},\varsigma \in \mathfrak{B}$ and $\tilde{\mathfrak{t}},\tilde{\mathfrak{s}}\in (0,1]$.

Definition 22.

An $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$ if it satisfies the following conditions:

(i) $((\theta -(\vartheta -\phi )),\tilde{\mathfrak{t}}),(\vartheta ,\tilde{\mathfrak{s}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow (\theta -\phi ,\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}})\acute{\in }\vee {\acute{q}}_{\check{k}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$;

(ii) $((\theta -(\vartheta -\phi )),\tilde{\mathfrak{t}}),(\vartheta ,\tilde{\mathfrak{s}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow (\theta -\phi ,\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}})\acute{\in }\vee {\acute{q}}_{\check{k}}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$,

for every $\theta ,\vartheta ,\phi \in \mathfrak{Y},\varsigma \in \mathfrak{B}$ and $\tilde{\mathfrak{t}},\tilde{\mathfrak{s}}\in (0,1]$.

Theorem 8.

An $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$ if and only if it satisfies:

(i) ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2$;

(ii) ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge [1-\check{k}]/2$;

for any $\theta ,\vartheta ,\phi \in \mathfrak{Y}$, and $\varsigma \in \mathfrak{B}$.

Proof. 

(i) First, let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$.

Case 1: Let ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )>[1-\check{k}]/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2={\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. If possible, let ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. Then let $\tilde{\mathfrak{t}}$ be a real number such that ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )>\tilde{\mathfrak{t}}>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))<\tilde{\mathfrak{t}},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )<\tilde{\mathfrak{t}}$, for some $\tilde{\mathfrak{t}}\in (0,1)$, i.e., $(\theta -(\vartheta -\phi ),\tilde{\mathfrak{t}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},$$(\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. However, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )>\tilde{\mathfrak{t}}$. Then $(\theta -\vartheta ,\tilde{\mathfrak{t}})\overline{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}>2\tilde{\mathfrak{t}}$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}>2({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta ))>2\times [1-\check{k}]/2=1-\check{k}$, i.e., ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}<1-\check{k}$. Therefore, $(\theta -\vartheta ,\tilde{\mathfrak{t}})\overline{{\acute{q}}_{\check{k}}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Hence, $(\theta -\vartheta ,\tilde{\mathfrak{t}})\overline{\acute{\in }\vee {\acute{q}}_{\check{k}}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$, which contradicts the fact that $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$ of $\mathfrak{Y}$. Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )={\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2$.

Case 2: Let ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le [1-\check{k}]/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi ))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )=[1-\check{k}]/2$. If possible, let ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2=[1-\check{k}]/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\le [1-\check{k}]/2$, and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le [1-\check{k}]/2$. So, $(\theta ,[1-\check{k}]/2)\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},(\vartheta ,[1-\check{k}]/2)\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. However, $(\theta -\vartheta ,[1-\check{k}]/2)\overline{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+[1-\check{k}]/2>[1-\check{k}]/2+[1-\check{k}]/2=1-\check{k}$ implies $(\theta -\vartheta ,[1-\check{k}]/2)\overline{{\acute{q}}_{\check{k}}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Hence, $(\theta -\vartheta ,[1-\check{k}]/2)\overline{\acute{\in }\vee {\acute{q}}_{\check{k}}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$, which again fails due to the fact that ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$. Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le [1-\check{k}]/2={\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2$.

Converse part:

Let

$${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2.$$

(23)

Let $\theta ,\vartheta \in \mathfrak{Y},\varsigma \in \mathfrak{B}$ and $\tilde{\mathfrak{t}},\tilde{\mathfrak{s}}\in (0,1]$ s.t $(\theta -(\vartheta -\phi ),\tilde{\mathfrak{t}}),(\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\le \tilde{\mathfrak{t}}$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{s}}$. Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$. By (23), we have ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}\vee [1-\check{k}]/2$.

Let $[1-\check{k}]/2\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$. Then $[1-\check{k}]/2\vee \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}=\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$. Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$.

$$(\theta -\vartheta ,\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(24)

Next, let $[1-\check{k}]/2<\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$. Then $[1-\check{k}]/2\vee \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}=[1-\check{k}]/2$. Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}\vee [1-\check{k}]/2=[1-\check{k}]/2$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}<[1-\check{k}]/2+[1-\check{k}]/2=1-\check{k}$.

$$(\theta -\vartheta ,\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}){\acute{q}}_{\check{k}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(25)

From (24) and (25), we have

$$(\theta -(\vartheta -\phi ),\tilde{\mathfrak{t}}),(\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow (\theta -\vartheta ,\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}})\acute{\in }\vee {\acute{q}}_{\check{k}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(26)

Therefore, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in }\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$.

(ii) First, let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $(\acute{\in }\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$.

Case 1: Let ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )<[1-\check{k}]/2$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge [1-\check{k}]/2={\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. If possible, let ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. Then let $\tilde{\mathfrak{t}}$ be a real number such that ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<\tilde{\mathfrak{t}}<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))>\tilde{\mathfrak{t}},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )>\tilde{\mathfrak{t}}$, for some $\tilde{\mathfrak{t}}\acute{\in }(0,1)$, i.e., $(\theta ,\tilde{\mathfrak{t}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]},(\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. However, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<\tilde{\mathfrak{t}}$. Then $(\theta -\vartheta ,\tilde{\mathfrak{t}})\overline{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}<2\tilde{\mathfrak{t}}$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}<2({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta ))<2\times [1-\check{k}]/2=1-\check{k}$, i.e., ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}<1-\check{k}$. Therefore, $(\theta -\vartheta ,\tilde{\mathfrak{t}})\overline{{\acute{q}}_{\check{k}}}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Hence, $(\theta -\vartheta ,\tilde{\mathfrak{t}})\overline{\acute{\in }\vee {\acute{q}}_{\check{k}}}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$, which contradicts the fact that $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$. Therefore, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )={\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge [1-\check{k}]/2$.

Case 2: Let ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge [1-\check{k}]/2$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )=[1-\check{k}]/2$. If possible, let ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge [1-\check{k}]/2=[1-\check{k}]/2$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\ge [1-\check{k}]/2$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge [1-\check{k}]/2$. Thus, $(\theta ,[1-\check{k}]/2)\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]},(\vartheta ,[1-\check{k}]/2)\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. However, $(\theta -\vartheta ,[1-\check{k}]/2)\overline{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+[1-\check{k}]/2<[1-\check{k}]/2+[1-\check{k}]/2=1-\check{k}$, i.e., $(\theta -\vartheta ,[1-\check{k}]/2)\overline{{\acute{q}}_{\check{k}}}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Hence, $(\theta -\vartheta ,[1-\check{k}]/2)\overline{\acute{\in }\vee {\acute{q}}_{\check{k}}}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$, which contradicts the fact that ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$. Therefore, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge [1-\check{k}]/2={\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge [1-\check{k}]/2$.

Converse part:

Let

$${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge [1-\check{k}]/2.$$

(27)

Let $\theta ,\vartheta \in \mathfrak{Y},\varsigma \in \mathfrak{B}$ and $\tilde{\mathfrak{t}},\tilde{\mathfrak{s}}\in (0,1]$ s.t $(\theta -(\vartheta -\phi ),\tilde{\mathfrak{t}}),(\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\ge \tilde{\mathfrak{t}}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge \tilde{\mathfrak{s}}$. Therefore, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$. By (27), we have ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}\wedge [1-\check{k}]/2$. Now, let $[1-\check{k}]/2\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$. Then $[1-\check{k}]/2\wedge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}=\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$,

$$(\theta -\vartheta ,\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(28)

Next, let $[1-\check{k}]/2<\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}$, $[1-\check{k}]/2\wedge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}=[1-\check{k}]/2$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}\wedge [1-\check{k}]/2=[1-\check{k}]/2$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}>[1-\check{k}]/2+[1-\check{k}]/2=1-\check{k}$,

$$(\theta -\vartheta ,\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}){\acute{q}}_{\check{k}}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(29)

From (28) and (29), we have

$$(\theta -(\vartheta -\phi ),\tilde{\mathfrak{t}}),(\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow (\theta -\vartheta ,\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}})\acute{\in }\vee {\acute{q}}_{\check{k}}{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(30)

Therefore, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$. Hence, (26) and (30) $\Rightarrow \tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$.

□

An $(\acute{\in },\acute{\in })$-$U{I}_{FS}{H}_{ID}$ is always an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$ of $\mathfrak{Y}$, but not conversely, which can be seen from the following example.

Example 4.

Consider a subtraction BG-algebra $\mathfrak{Y}=\{0,\theta ,\vartheta ,\phi \}$ with the following Cayley

Table 5. Example of $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$.

−	0	$\theta$	$\vartheta$	$\phi$
0	0	$\theta$	$\vartheta$	$\phi$
$\theta$	$\theta$	0	$\phi$	$\vartheta$
$\vartheta$	$\vartheta$	$\phi$	0	$\theta$
$\phi$	$\phi$	$\vartheta$	$\theta$	0

:

Define $\tilde{\mathfrak{N}}[\varsigma ]$ by

−	0	$\theta$	$\vartheta$	$\phi$
${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$	0.41	0.29	0.41	0.41
${\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}$	0.69	0.54	0.69	0.69

Then $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}{H}_{ID}$, but it is not an $(\acute{\in },\acute{\in })$-$U{I}_{FS}{H}_{ID}$, since ${\theta }_{0.6}\acute{\in }{\omega }_{\tilde{\mathfrak{N}}}[\varsigma ],{\vartheta }_{0.6}\acute{\in }{\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}$, but ${(\theta -\vartheta )}_{0.59}={\phi }_{0.59}\acute{\notin }{\omega }_{\tilde{\mathfrak{N}}[\varsigma ]}$.

Theorem 9.

An $I_{FS}SU$$\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$ and if ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))>[1-\check{k}]/2,{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))<[1-\check{k}]/2$, then $\tilde{\mathfrak{N}}[\varsigma ]$ is also an $(\acute{\in },\acute{\in })$-$U{I}_{FS}{H}_{ID}$.

Proof. 

Let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))>[1-\check{k}]/2$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))<[1-\check{k}]/2$. Let $(\theta -(\vartheta -\phi ),\tilde{\mathfrak{t}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},(\vartheta ,\tilde{\mathfrak{s}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then we have $[1-\check{k}]/2<{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\le \tilde{\mathfrak{t}}and[1-\check{k}]/2<{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{s}}$. Thus, $\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}>[1-\check{k}]/2$, i.e., ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )>[1-\check{k}]/2$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}>[1-\check{k}]/2+[1-\check{k}]/2=1-\check{k}$. Since ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}{H}_{ID}$, either ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}<1-\check{k}or{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}}\Rightarrow (\theta -\vartheta ,\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Therefore,

$$(\theta -(\vartheta -\phi ),\tilde{\mathfrak{t}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]},(\vartheta ,\tilde{\mathfrak{s}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow (\theta -\vartheta ,\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{s}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(31)

Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is $(\acute{\in },\acute{\in })$-$U{I}_{FS}{H}_{ID}$.

Again, let $(\theta -(\vartheta -\phi ),\tilde{\mathfrak{t}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]},(\vartheta ,\tilde{\mathfrak{s}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Then we have $t\le {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))<[1-\check{k}]/2$ and $\tilde{\mathfrak{s}}\le {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )<[1-\check{k}]/2$.

Thus, $\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}<[1-\check{k}]/2$, i.e., ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<[1-\check{k}]/2$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}<[1-\check{k}]/2+[1-\check{k}]/2=1-\check{k}$. Since ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}{H}_{ID}$, either ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}or{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+t\wedge \tilde{\mathfrak{s}}>1-\check{k}$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}}\Rightarrow (\theta -\vartheta ,\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Therefore,

$$(\theta -(\vartheta -\phi ),\tilde{\mathfrak{t}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]},(\vartheta ,\tilde{\mathfrak{s}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow (\theta -\vartheta ,\tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{s}})\acute{\in }{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}.$$

(32)

Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is $(\acute{\in },\acute{\in })$-$U{I}_{FS}{H}_{ID}$. Hence, (31) and (32) $\Rightarrow \tilde{\mathfrak{N}}[\varsigma ]$ is also an $(\acute{\in },\acute{\in })$-$U{I}_{FS}{H}_{ID}$. □

Theorem 10.

Let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $I_{F}S$ if and only if $U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}}):=\{\theta \acute{\in }\mathfrak{Y}:{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\le \tilde{\mathfrak{t}}and{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}\ge \tilde{\mathfrak{t}}\}$ is a h-ideal of $\mathfrak{Y}$ for all $\tilde{\mathfrak{t}}\in (0,[1-\check{k}]/2)$.

Proof. 

Suppose that $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$.

(i) Let $\tilde{\mathfrak{t}}\in (0,[1-\check{k}]/2)$ and $\theta -(\vartheta -\phi ),\vartheta \acute{\in }U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\le \tilde{\mathfrak{t}}$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\le \tilde{\mathfrak{t}}$. It follows from (23) that ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{t}}\vee [1-\check{k}]/2=\tilde{\mathfrak{t}}$. Thus, $\theta -\vartheta \acute{\in }U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$. Thus, $U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$ is an h-ideal of $\mathfrak{Y}$.

Conversely, suppose that $U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$ is an h-ideal of $\mathfrak{Y}$, for all $\tilde{\mathfrak{t}}\in (0,[1-\check{k}]/2)$. If (23) is not true, then there exist $\theta -(\vartheta -\phi ),\vartheta \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$ such that ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2$. Hence, we can see that $\tilde{\mathfrak{t}}\in (0,1)$ such that ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )>\tilde{\mathfrak{t}}\ge {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2$. Then $\tilde{\mathfrak{t}}\acute{\in }(0,[1-\check{k}]/2)$ and $\theta -(\vartheta -\phi ),\vartheta \acute{\in }U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$. In addition, $U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$ is an h-ideal of $\mathfrak{Y}$ which implies $\theta -\vartheta \acute{\in }U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge \tilde{\mathfrak{t}}$, which is a contradiction. Therefore, (23) is valid, and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$ of $\mathfrak{Y}$.

(ii) Let $\tilde{\mathfrak{t}}\in (0,[1-\check{k}]/2)and\theta -(\vartheta -\phi ),\vartheta \acute{\in }U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$. Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\ge \tilde{\mathfrak{t}}and{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\ge \tilde{\mathfrak{t}}\}$. It follows from (27) that ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge \tilde{\mathfrak{t}}\wedge \tilde{\mathfrak{t}}\wedge [1-\check{k}]/2=\tilde{\mathfrak{t}}$. Thus, $\theta -\vartheta \acute{\in }U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$. Thus $U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$ is an h-ideal of $\mathfrak{Y}$.

Conversely, suppose that $U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$ is an h-ideal of $\mathfrak{Y}$ for all $\tilde{\mathfrak{t}}\in (0,[1-\check{k}]/2)$. If (23) is not true, then there exists $\theta -(\vartheta -\phi ),\vartheta \acute{\in }\mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$ such that ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge [1-\check{k}]/2$. Hence, we can see that $\tilde{\mathfrak{t}}\in (0,1)$ such that ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<\tilde{\mathfrak{t}}\le {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\wedge [1-\check{k}]/2$. Thus, $\tilde{\mathfrak{t}}\in (0,[1-\check{k}]/2)$ and $\theta ,\vartheta \acute{\in }U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$. In addition, $U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$ is an h-ideal of $\mathfrak{Y}$ which implies that $\theta -\vartheta \acute{\in }U(\tilde{\mathfrak{N}}[\varsigma ];\tilde{\mathfrak{t}})$. Thus, ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le \tilde{\mathfrak{t}}$, which is a contradiction. Therefore, (27) is valid, and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$ of $\mathfrak{Y}$.

□

Theorem 11.

Let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $I_{FS}SU$. Then $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$ if and only if ${[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$, and ${[{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$ is an h-ideal of $\mathfrak{Y}$, for all $\tilde{\mathfrak{t}}\acute{\in }(0,1]$. We call ${[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$, and ${[{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$$(\acute{\in }\vee {\acute{q}}_{\check{k}})$-level h-ideals of $\mathfrak{Y}$.

Proof. 

Suppose that $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$ -$U{I}_{FS}{H}_{ID}$. To prove ${[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$ and ${[{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$ is a h-ideal of $\mathfrak{Y}$, let $\theta -(\vartheta -\phi ),\vartheta \acute{\in }{[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$, for $\tilde{\mathfrak{t}}\in (0,1]$. Then $(\theta -(\vartheta -\phi ),\tilde{\mathfrak{t}}),(\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{\acute{q}}_{\check{k}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$ implies ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\le \tilde{\mathfrak{t}}$ and ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{t}}$. Thus, $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_k)$-$U{I}_{FS}{H}_{ID}$, i.e., ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2$, for any $\theta ,\vartheta ,\phi \in \mathfrak{Y}$ and $\varsigma \in \mathfrak{B}$.

Now we have the following cases.

Case 1: Let ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\le \tilde{\mathfrak{t}},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{t}},$ and let $\tilde{\mathfrak{t}}<[1-\check{k}]/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2=\tilde{\mathfrak{t}}\vee \tilde{\mathfrak{t}}\vee [1-\check{k}]/2=[1-\check{k}]/2$. It follows that ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le [1-\check{k}]/2$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}<[1-\check{k}]/2+[1-\check{k}]/2=1-\check{k}$, i.e., $(\theta -\vartheta ,\tilde{\mathfrak{t}}){\acute{q}}_{\check{k}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$.

Next, let $\tilde{\mathfrak{t}}\ge [1-\check{k}]/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2\le \tilde{\mathfrak{t}}\vee \tilde{\mathfrak{t}}\vee [1-\check{k}]/2=\tilde{\mathfrak{t}}$. It follows that $(\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Thus, $(\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }\vee {\acute{q}}_{\check{k}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$, i.e., $(\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$.

Case 2: Let ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\le \tilde{\mathfrak{t}},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )+\tilde{\mathfrak{t}}<1-\check{k},$ and let $\tilde{\mathfrak{t}}<[1-\check{k}]/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2<\tilde{\mathfrak{t}}\vee (1-\check{k}-\tilde{\mathfrak{t}})\vee [1-\check{k}]/2=1-\check{k}-\tilde{\mathfrak{t}}$. It follows that ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<1-\check{k}-\tilde{\mathfrak{t}}$. So, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}<1-\check{k}$, i.e., $(\theta -\vartheta ,\tilde{\mathfrak{t}}){\acute{q}}_{\check{k}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$.

Next, let $\tilde{\mathfrak{t}}\ge [1-\check{k}]/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2\le \tilde{\mathfrak{t}}\vee (1-\check{k}-\tilde{\mathfrak{t}})\vee [1-\check{k}]/2=t$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le \tilde{\mathfrak{t}}$, i.e., $(\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Hence, $(\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }\vee \acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow (\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$.

Case 3: Let ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))+\tilde{\mathfrak{t}}<1-\check{k},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\le \tilde{\mathfrak{t}},$ and let $\tilde{\mathfrak{t}}<[1-\check{k}]/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2<(1-\check{k}-\tilde{\mathfrak{t}})\vee \tilde{\mathfrak{t}}\vee [1-\check{k}]/2=1-\check{k}-\tilde{\mathfrak{t}}$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<1-\check{k}-\tilde{\mathfrak{t}}\Rightarrow {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}<1-\check{k}$, i.e., $(\theta -\vartheta ,\tilde{\mathfrak{t}}){\acute{q}}_{\check{k}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$.

Next, let $\tilde{\mathfrak{t}}\ge [1-\check{k}]/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2=(1-\check{k}-\tilde{\mathfrak{t}})\vee \tilde{\mathfrak{t}}\vee [1-\check{k}]/2=\tilde{\mathfrak{t}}$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le \tilde{\mathfrak{t}}$, i.e., $(\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Hence, $(\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }\vee \acute{q}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow (\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$.

Case 4: ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))+\tilde{\mathfrak{t}}<1-\check{k},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )+\tilde{\mathfrak{t}}<1-\check{k},$ and let $\tilde{\mathfrak{t}}<[1-\check{k}]/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2<(1-\check{k}-\tilde{\mathfrak{t}})\vee (1-\check{k}-\tilde{\mathfrak{t}})\vee [1-\check{k}]/2=1-\check{k}-\tilde{\mathfrak{t}}$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<1-\check{k}-\tilde{\mathfrak{t}}$, i.e., ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}<1-\check{k}$. Hence, $(\theta -\vartheta ,\tilde{\mathfrak{t}}){\acute{q}}_{\check{k}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$.

Next, let $\tilde{\mathfrak{t}}\ge [1-\check{k}]/2$. Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\vartheta )\vee [1-\check{k}]/2=(1-\check{k}-\tilde{\mathfrak{t}})\vee (1-\check{k}-\tilde{\mathfrak{t}})\vee [1-\check{k}]/2=[1-\check{k}]/2\le \tilde{\mathfrak{t}}$. Thus, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le \tilde{\mathfrak{t}}$, i.e., $(\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}$. Hence, $(\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }\vee {\acute{q}}_{\check{k}}{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}\Rightarrow (\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$. Therefore, $\theta -(\vartheta -\phi ),\vartheta \acute{\in }{[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}\Rightarrow (\theta -\vartheta ,\tilde{\mathfrak{t}})\acute{\in }{[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$ is an h-ideal of $\mathfrak{Y}$. Similarly, we can prove ${[{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$ is a h-ideal of $\mathfrak{Y}$.

Conversely, let $\tilde{\mathfrak{N}}[\varsigma ]$ be an $U{I}_{FS}{H}_{ID}$ s.t ${[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$ and ${[{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$ is an h-ideal of $\mathfrak{Y}$ for all $\tilde{\mathfrak{t}}\in (0,1]$. To show $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$. Suppose $\tilde{\mathfrak{N}}[\varsigma ]$ is not an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$. Then there exists $\theta ,\vartheta ,\phi \in \mathfrak{Y}$ s.t at least one of ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\vee [1-\check{k}]/2$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\wedge [1-\check{k}]/2$ holds.

Suppose ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\vee [1-\check{k}]/2$ holds, then select $\tilde{\mathfrak{t}}\in (0,1]$ s.t

$${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )>\tilde{\mathfrak{t}}>{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\vee [1-\check{k}]/2.$$

(33)

Then ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))<\tilde{\mathfrak{t}},{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))<\tilde{\mathfrak{t}}\Rightarrow \theta -(\vartheta -\phi ),\vartheta \acute{\in }{({\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}\subset {[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$ is an h-ideal. Therefore, $(\theta -\phi )\acute{\in }{[{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}\Rightarrow {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le \tilde{\mathfrak{t}}or{\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}<1-\check{k}$, which contradicts (33).

Next, let ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\wedge [1-\check{k}]/2$ hold, then select $\tilde{\mathfrak{t}}\in (0,1]$ s.t

$${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )<\tilde{\mathfrak{t}}<{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))\wedge [1-\check{k}]/2.$$

(34)

Then ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))>\tilde{\mathfrak{t}},{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -(\vartheta -\phi ))>\tilde{\mathfrak{t}}\Rightarrow \theta -(\vartheta -\phi ),\vartheta \acute{\in }{({\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]})}_{\tilde{\mathfrak{t}}}\subset {[{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}$ is an h-ideal. Therefore, $(\theta -\phi )\acute{\in }{[{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}]}_{\tilde{\mathfrak{t}}}\Rightarrow {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge \tilde{\mathfrak{t}}or{\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )+\tilde{\mathfrak{t}}>1-\check{k}$, which contradicts (34). Hence, ${\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\le {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee {\varkappa }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\vee [1-\check{k}]/2$ and ${\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}(\theta -\phi )\ge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge {\varpi }_{\tilde{\mathfrak{N}}[\varsigma ]}((\theta -(\vartheta -\phi )))\wedge [1-\check{k}]/2$. Thus, $\tilde{\mathfrak{N}}[\varsigma ]$ is an $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$. □

6. Conclusions

In this paper, we introduced the notions of $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SAs$ in subtraction BG-algebras. We provided different characterizations and some equivalent conditions of $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}SAs$ in terms of the level subsets of subtraction BG-algebras. It was shown that $(\acute{q},\acute{q})$-$U{I}_{FS}SA$ are $(\acute{\in },\acute{\in })$-$U{I}_{FS}SA$ but the converse does not hold and an example is provided. We introduced $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}IDs$ and some of its usual properties. Also, $h^{-1}(\tilde{\mathfrak{N}}[\varsigma ])$ is an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$. Moreover, if $h^{-1}(\tilde{\mathfrak{N}}[\varsigma ])$ are an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$, then $\tilde{\mathfrak{N}}[\varsigma ]$ are an $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}ID$. Finally, we characterized $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$U{I}_{FS}{H}_{ID}$ which is a generalization of $(\acute{\in },\acute{\in }\vee \acute{q})$-$U{I}_{FS}{H}_{ID}$. In the future, we may carry out $(\acute{\in },\acute{\in }\vee {\acute{q}}_{\check{k}})$-$BU{I}_{FS}SAs$ and its applications in decision-making problems.