Third Zadeh’s Intuitionistic Fuzzy Implication †

: George Klir and Bo Yuan named after Lotﬁ Zadeh the implication p → q = max ( 1 − p ,min ( p , q )) (also Early Zadeh implication ). In a series of papers, the author introduced two intuitionistic fuzzy forms of Zadeh’s implication and studied their basic properties. In the present paper, a new (third) intuitionistic fuzzy form of Zadeh’s implication is proposed and some of its properties are studied.


Introduction
In the present article, a new operation "implication" over intuitionistic fuzzy sets is introduced. It is based on the definition of the fuzzy implication, proposed by George Klir and Bo Yuan [1] and named after Lotfi Zadeh (also Early Zadeh implication), which has the form p → q = max(1 − p, min(p, q)) see also [2], as well as on the two previous intuitionistic fuzzy implications introduced by the author [3,4] also inspired by the Zadeh implication.
In the beginning, the necessary concepts from intuitionistic fuzzy set theory will be given.
For two IFSs A and B: Therefore, for every two IFSs A and B: An IFS A is called intuitionistic fuzzy tautological set (IFTS) if and only if (iff) for every and it is called tautological set iff for every x ∈ E: µ A (X) = 1, ν A (x) = 0. For two IFSs A and B: Therefore, for every two IFSs A and B: and

Let us have two IFSs
Now, we introduce the new (third) Zadeh's intuitionistic fuzzy implication with the form: First, we check that the definition is correct. Obviously, the membership part and the non-membership part We must check the following cases.
(by the assumption in 1.1).
Mathematics 2021, 9, 619 Therefore, the operation is defined correctly. Now, we can see that there is not a relation between the third Zadeh's implication and each one of the first two Zadeh's implications. Really, if the universe is E = {x} and i.e., both sets are not comparable.
The new implication generates the classical negation of IFS A, because [5,6]). We can check directly that In the particular case, we have Four different geometrical interpretations of the element x ∈ E in IFSs A and B, i.e., with degrees µ A (x) and ν A (x), and µ B (x) and ν B (x); and the element x from IFS A → Z, 3  A → Z,3 E * = { x, max(1, min(ν A (x), 0)), min(0, max(µ A (x), 1)) |x ∈ E} In the particular case, we have @ @ @ @ @ @ @ @ @ @ @ @ (0,1)     Following [6], we will mention that if the axiom is valid as an intuitionistic fuzzy tautology (IFT), that axiom is marked with an asterisk ( * ). Such are the axioms:     N(x))), where N is an operation for a negation. Axiom 9. I is a continuous function.
Following [6], we will mention that if the axiom is valid as an intuitionistic fuzzy tautology (IFT), that axiom is marked with an asterisk ( * ). Such are the axioms:  Fig. 3: Geometrical interpretations of elements x A , x B and x A→ Z,3 B -the second scenario Fig. 4: Geometrical interpretations of elements x A , x B and x A→ Z,3 B -the third scenario 49 @ @ @ @ @ @ @ @ @ @ @ @ (0,1)        N(x))), where N is an operation for a negation. Axiom 9. I is a continuous function. Following [6], we will mention that if the axiom is valid as an intuitionistic fuzzy tautology (IFT), that axiom is marked with an asterisk ( * ). Such are the axioms: Proof. Let A, B and C be three IFSs. In [5,6], the relation ⊆ is defined by Now, we see that for each x ∈ E: (from µ A (x) ≤ µ B (x)). Therefore, Axiom 1 is valid. From E * → Z,3 A = A that we checked above, it follows that Axiom 4 is valid, too.
Since functions max and min are continuous, Axiom 9 is valid. For the rest axioms we can construct counterexamples. For example, for the universe E = {x} and the IFSs Therefore, Axiom 2 is not valid. In the proofs of Theorem 2, we will construct two other counterexamples.
Theorem 2. Implication → Z,3 satisfies Axioms 3* and 5*, but it does not satisfy Axioms 3 and 5, i.e., in the forms O * → Z,3 A and A → Z,3 A are IFTSs, but O * → Z,3 A = E * and A → Z,3 A = E * are not tautological sets.
To the proofs of both theorems, we can add also that Axiom 7* is not valid (and hence, Axiom 7, too), because the counterexample with the universe E = {x} and the IFSs A = { x, 1 2 , 1 2 }, B = { x, 0, 1 }. In this case B ⊂ A (strong inclusion), while