On the Classification of BolMoufang Type of Some Varieties of Quasi Neutrosophic Triplet Loop (Fenyves BCIAlgebras)
Abstract
:1. Introduction
1.1. BCIalgebra, Quasigroups, Loops and the Fenyves Identities
 1.
 $\left(\right(x*y)*(x*z\left)\right)*(z*y)=0$;
 2.
 $x*0=x$;
 3.
 $x*y=0$ and $y*x=0$ ⟹ $x=y$.
 1.
 $\left(\right(x*y)*(x*z\left)\right)*(z*y)=0$;
 2.
 $(x*(x*y\left)\right)*y=0$;
 3.
 $x*x=0$;
 4.
 $x*y=0$ and $y*x=0$ imply $x=y$.


 1.
 X is associative.
 2.
 $0*x=x$.
 3.
 $x*y=y*x\phantom{\rule{3.33333pt}{0ex}}\forall \phantom{\rule{3.33333pt}{0ex}}x,y\in X$.
 1.
 X is psemisimple
 2.
 $(x*y)*(z*u)=(x*z)*(y*u)$.
 3.
 $0*(y*x)=x*y$.
 4.
 $(x*y)*(x*z)=z*y$.
 5.
 $z*x=z*y$ implies $x=y$. (the left cancellation law i.e., LCL)
 6.
 $x*y=0$ implies $x=y$.
 1.
 X is quasiassociative.
 2.
 $x*(0*y)=0$ implies $x*y=0$.
 3.
 $0*x=0*(0*x)$.
 4.
 $(0*x)*x=0$.
 1.
 $(x*y)*(x*z)\u2a7dz*y$;
 2.
 $x*(x*y)\u2a7dy$;
 3.
 $x*y=0$ if and only if $x\u2a7dy$.
 1.
 $x*z=y*z$ implies $x=y$. (the right cancellation law i.e., RCL)
 2.
 $(y*x)*(z*x)=y*z$.
 3.
 $(x*y)*(x*z)=0*(y*z)$.
 1.
 $x*(0*y)=y$.
 2.
 $0*x=0\Rightarrow x=0$.
1.2. BCIAlgebras as a Quasi Neutrosophic Triplet Loop
 1.
 If there exist $b,c\in X$ such that $a*b=a$ and $a*c=b$, then a is called an NTelement with (rr)property. If every $a\in X$ is an NTelement with (rr)property, then, $(X,*)$ is called a (rr)quasi NTL.
 2.
 If there exist $b,c\in X$ such that $a*b=a$ and $c*a=b$, then a is called an NTelement with (rl)property. If every $a\in X$ is an NTelement with (rl)property, then, $(X,*)$ is called a (rl)quasi NTL.
 3.
 If there exist $b,c\in X$ such that $b*a=a$ and $c*a=b$, then a is called an NTelement with (ll)property. If every $a\in X$ is an NTelement with (ll)property, then, $(X,*)$ is called a (ll)quasi NTL.
 4.
 If there exist $b,c\in X$ such that $b*a=a$ and $a*c=b$, then a is called an NTelement with (lr)property. If every $a\in X$ is an NTelement with (lr)property, then, $(X,*)$ is called a (lr)quasi NTL.
 5.
 If there exist $b,c\in X$ such that $a*b=b*a=a$ and $a*c=b$, then a is called an NTelement with (lrr)property. If every $a\in X$ is an NTelement with (lrr)property, then, $(X,*)$ is called a (lrr)quasi NTL.
 6.
 If there exist $b,c\in X$ such that $a*b=b*a=a$ and $c*a=b$, then a is called an NTelement with (lrl)property. If every $a\in X$ is an NTelement with (lrl)property, then, $(X,*)$ is called a (lrl)quasi NTL.
 7.
 If there exist $b,c\in X$ such that $a*b=a$ and $a*c=c*a=b$, then a is called an NTelement with (rlr)property. If every $a\in X$ is an NTelement with (rlr) property, then, $(X,*)$ is called a (rlr)quasi NTL.
 8.
 If there exist $b,c\in X$ such that $b*a=a$ and $a*c=c*a=b$, then a is called an NTelement with (llr)property. If every $a\in X$ is an NTelement with (llr)property, then, $(X,*)$ is called a (llr)quasi NTL.
 9.
 If there exist $b,c\in X$ such that $a*b=b*a=a$ and $a*c=c*a=b$, then a is called an NTelement with (lrlr)property. If every $a\in X$ is an NTelement with (lrlr)property, then, $(X,*)$ is called a (lrlr)quasi NTL.
2. Main Results
 1.
 A BCI algebra X is a quasigroup if and only if it is psemisimple.
 2.
 A BCI algebra X is a loop if and only if it is associative.
 3.
 An associative BCI algebra X is a Boolean group.
 From Theorem 7 and Theorem 4, psemisimplicity is equivalent to the left and right cancellation laws, which consequently implies that X is a quasigroup if and only if it is psemisimple.
 One of the axioms that a BCIalgebra satisfies is $x*0=x$ for all $x\in X$. So, 0 is already the right identity element. Now, from Theorem 3, associativity is equivalent to $0*x=x$ for all $x\in X$. So, 0 is also the left identity element of X. The conclusion follows.
 In a BCIalgebra, $x*x=0$ for all $x\in X$. And 0 is the identity element of X. Hence, every element is the inverse of itself.
 1.
 $0\in {N}_{\rho}\left(X\right)$.
 2.
 $0\in {N}_{\lambda}\left(X\right),{N}_{\mu}\left(X\right)$ implies X is quasiassociative.
 3.
 If $0\in {N}_{\lambda}\left(X\right)$, then the following are equivalent:
 (a)
 X is psemisimple.
 (b)
 $xy=0y\xb7x$ for all $x,y\in L$.
 (c)
 $xy=0x\xb7y$ for all $x,y\in L$.
 4.
 If $0\in {N}_{\lambda}\left(X\right)$ or $0\in {N}_{\mu}\left(X\right)$, then X is psemisimple if and only if X is associative.
 5.
 If $0\in N\left(X\right)$, then X is psemisimple if and only if X is associative.
 6.
 If $(X,*,0)$ is a BCKalgebra, then
 (a)
 $0\in {N}_{\lambda}\left(X\right)$.
 (b)
 $0\in {N}_{\mu}\left(X\right)$ implies X is a trivial BCKalgebra.
 7.
 The following are equivalent:
 (a)
 X is associative.
 (b)
 $x\in {N}_{\lambda}\left(X\right)$ for all $x\in X$.
 (c)
 $x\in {N}_{\rho}\left(X\right)$ for all $x\in X$.
 (d)
 $x\in {N}_{\mu}\left(X\right)$ for all $x\in X$.
 (e)
 $0\in C\left(X\right)$.
 (f)
 $x\in C\left(X\right)$ for all $x\in X$.
 (g)
 $x\in Z\left(X\right)$ for all $x\in X$.
 (h)
 $0\in Z\left(X\right)$.
 (i)
 X is a (lrr)QNTL.
 (j)
 X is a (lrl)QNTL.
 (k)
 X is a (lrlr)QNTL
 8.
 If $(X,*,0)$ is a BCKalgebra and $0\in C\left(X\right)$, then X is a trivial BCKalgebra.





 
 
 
 

 Let X be an ${F}_{1}$algebra. Then $xy*zx=(xy*z)x$. With $z=y$, we have $xy*yx=(xy*y)x$ which implies $xy*yx=(xy*x)y=(xx*y)y=(0*y)y=0*(y*y)$ (since $0\in {N}_{\lambda}\left(X\right)$; this is achieved by putting $y=x$ in the ${F}_{1}$ identity) $=0*0=0$. This implies $xy*yx=0$. Now replacing x with y, and y with x in the last equation gives $yx*xy=0$ implying that $x*y=y*x$ as required.
 Let X be an ${F}_{2}$algebra. Then $xy*zx=(x*yz)x$. With $y=z$, we have $xz*zx=(x*zz)x=(x*0)*x=x*x=0$ implying that $xz*zx=0$. Now replacing x with z, and z with x in the last equation gives $zx*xz=0$ implying that $x*z=z*x$ as required.
 Let X be a ${F}_{4}$algebra. Then, $xy*zx=x(yz*x)$. Put $y=x$ and $z=0$, then you get $0*0x=x$ which means X is psemisimple. Put $x=0$ and $y=0$ to get $0z=0*0z$ which implies that X is quasiassociative (Theorem 5). Thus, by Theorem 9, X is associative.
 Let X be an ${F}_{6}$algebra. Then, $(xy*z)x=x(y*zx)$. Put $x=y=0$ to get $0z=0*0z$ which implies that X is quasiassociative (Theorem 5). Put $y=0$ and $z=x$, then we have $0*x=x$. Thus, X is associative.
 Let X be an ${F}_{7}$algebra. Then $(xy*z)x=x(yz*x)$. With $z=0$, we have $xy*x=x(y*x)$. Put $y=x$ in the last equation to get $xx*x=(x*xx)$ implying $0*x=x$.
 Let X be an ${F}_{9}$algebra. Then $(x*yz)x=x(yz*x)$. With $z=0$, we have $(x*y)*x=x(y*x)$. Put $y=x$ in the last equation to get $(x*x)x=x(x*x)$ implying $0*x=x$.
 Let X be an ${F}_{10}$algebra. Then, $x(y*zx)=x(yz*x)$. Put $y=x=z$, then we have $x*0x=0$. So, $0x=0\Rightarrow x=0$. which means that X is psemisimple (Theorem 8(2)). Hence, X has the LCL by Theorem 4. Thence, the ${F}_{10}$ identity $x(y*zx)=x(yz*x)\Rightarrow y*zx=yz*x$ which means that X is associative.
 Let X be an ${F}_{11}$algebra. Then $xy*xz=(xy*x)z$. With $y=0$, we have $x*xz=xx*z$. Put $z=x$ in the last equation to get $x=0*x$ as required.
 Let X be an ${F}_{12}$algebra. Then $xy*xz=(x*yx)z$. With $z=0$, we have $xy*x=x*yx$. Put $y=x$ in the last equation to get $xx*x=x*xx$ implying $0*x=x$ as required.
 Let X be an ${F}_{13}$algebra. Then $xy*xz=x(yx*z)$. With $z=0$, we have $(x*y)x=x*yx$ which implies $(x*x)y=x*yx$ which implies $0*y=x*yx$. Put $y=x$ in the last equation to get $0*x=x$ as required.
 Let X be an ${F}_{14}$algebra. Then $xy*xz=x(y*xz)$. With $z=0$, we have $xy*x=x*yx$. Put $y=x$ in the last equation to get $0*x=x$ as required.
 Let X be an ${F}_{15}$algebra. Then $(xy*x)z=(x*yx)z$. With $z=0$, we have $(xy*x)=(x*yx)$. Put $y=x$ in the last equation to get $0*x=x$ as required.
 Let X be an ${F}_{16}$algebra. Then $(xy*x)z=x(yx*z)$. With $z=0$, we have $(xy*x)=(x*yx)$. Put $y=x$ in the last equation to get $0*x=x$ as required.
 Let X be an ${F}_{17}$algebra. Then $(xy*x)z=x(y*xz)$. With $z=0$, we have $(xy*x)=x(y*x)$. Put $y=x$ in the last equation to get $0*x=x$ as required.
 Let X be an ${F}_{18}$algebra. Then $(x*yx)z=x(yx*z)$. With $y=0$, we have $(x*0x)z=x(0x*z)$. Since $0\in {N}_{\lambda}\left(X\right)$ and $0\in {N}_{\mu}\left(X\right)$, (these are obtained by putting $x=0$ and $x=y$ respectively in the ${F}_{18}$identity), the last equation becomes $(x0*x)z=x(0*xz)=x0*xz=x*xz$ which implies $0*z=x*xz$. Put $x=z$ in the last equation to get $0*z=z$ as required.
 This is similar to the proof for ${F}_{10}$algebra.
 Let X be an ${F}_{22}$algebra. Then $yx*zx=(y*xz)x$. Put $y=x,z=0$, then $0x=0*0x$ which implies that X is quasiassociative. By Theorem 10, the ${F}_{22}$ identity implies that $yx*zx=yx*xz$. Substitute $x=0$ to get $yz=y*0z$. Now, put $y=z$ in this to get $z*0z=0$. So, $0z=0\Rightarrow z=0$. Hence, X is psemisimple (Theorem 8(2)). Thus, by Theorem 9, X is associative.
 Let X be an ${F}_{23}$algebra. Then $yx*zx=y(xz*x)$. With $z=0$, we have $yx*0x=y(x*x)$ which implies $yx*0x=y$. Since $0\in {N}_{\mu}\left(X\right)$, (this is obtained by putting $z=x$ in the ${F}_{23}$identity), the last equation becomes $(yx*0)*x=y$ which implies $(yx*x)=y$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{24}$algebra. Then $yx*zx=y(x*zx)$. With $z=0$, we have $yx*0x=y(x*0x)$. Since $0\in {N}_{\mu}\left(X\right)$,(this is obtained by putting $x=0$ in the ${F}_{24}$identity), the last equation becomes $\left(\right(yx)0*x)=y(x0*x)$ which implies $yx*x=y$. Put $y=x$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{25}$algebra. Then $(yx*z)x=(y*xz)x$. Put $x=0$, then $yz=y*0z$. Substitute $z=y$, then $y*0y=0$. So, $0y=0\Rightarrow y=0$. Hence, X is psemisimple (Theorem 8(2)). Hence, X has the RCL by Theorem 7. Thence, the ${F}_{25}$ identity $(yx*z)x=(y*xz)x$ implies $yx*z=y*xz$. Thus, X is associative.
 Let X be an ${F}_{26}$algebra. Then $(yx*z)x=y(xz*x)$. With $z=0$, we have $yx*x=y$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{27}$algebra. Then $(yx*z)x=y(x*zx)$. Put $z=x=y$, then $0x*x=0$ which implies X is quasiassociative. Put $x=0$ and $y=z$ to get $z*0z=0$. So, $0z=0\Rightarrow z=0$. Hence, X is psemisimple (Theorem 8(2)). Thus, by Theorem 9, X is associative.
 Let X be an ${F}_{28}$algebra. Then $(y*xz)x=y(xz*x)$. With $z=0$, we have $yx*x=y$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 The proof of this is similar to the proof for ${F}_{10}$algebra.
 Let X be an ${F}_{31}$algebra. Then $yx*xz=(yx*x)z$. By Theorem 10, the ${F}_{31}$ identity becomes ${F}_{25}$ identity which implies that X is associative.
 Let X be an ${F}_{32}$algebra. Then $yx*xz=(y*xx)z$. With $z=0$, we have $yx*x=y$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{33}$algebra. Then $yx*xz=y(xx*z)$. With $z=0$, we have $yx*x=y$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{34}$algebra. Then $yx*xz=y(x*xz)$. With $z=0$, we have $yx*x=y$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{35}$algebra. Then $(yx*x)z=(y*xx)z$. With $z=0$, we have $yx*x=y$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{36}$algebra. Then $(yx*x)z=y(xx*z)$. With $z=0$, we have $yx*x=y$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{37}$algebra. Then $(yx*x)z=y(x*xz)$. With $z=0$, we have $yx*x=y$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{38}$algebra. Then, $yz=y*0z$. Put $z=y$, then $y*0y=0$. So, $0y=0\Rightarrow y=0$. Hence, X is psemisimple (Theorem 8(2)). Now, put $y=x$, then $xz=x*0z$. Now, substitute $x=0$ to get $0z=0*0z$ which means that X is quasiassociative. Thus, by Theorem 9, X is associative.
 Let X be an ${F}_{40}$algebra. By the ${F}_{40}$ identity, $y*0z=y(x*xz)$. Put $z=x=y$ to get $0*0x=0$. So, $0x=0\Rightarrow x=0$. Hence, X is psemisimple (Theorem 8(2)). Thus, X has the LCL by Theorem 4. Thence, the ${F}_{40}$ identity $y(xx*z)=y(x*xz)$ becomes $0*z=x*xz$. Substituting $z=x$, we get $0x=x$ which means that X is associative.
 Let X be an ${F}_{41}$algebra. Then $xx*yz=(x*xy)z$. With $z=0$, we have $0*y=x*xy$. Put $y=x$ in the last equation to get $0*x=x$ as required.
 Let X be an ${F}_{43}$algebra. Then $xx*yz=x(x*yz)$. With $z=0$, we have $0*y=x(x*y)$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{44}$algebra. Then $xx*yz=x(xy*z)$. With $z=0$, we have $0*y=x(x*y)$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{45}$algebra. Then $(x*xy)z=(xx*y)z$. With $z=0$, we have $x*xy=0*y$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{47}$algebra. Then $(x*xy)z=x(xy*z)$. With $y=0$, we have $0*z=x(x*z)$. Put $x=z$ in the last equation to get $0*z=z$ as required.
 Let X be an ${F}_{48}$algebra. Then $(xx*y)z=x(x*yz)$. With $z=0$, we have $0*y=x*xy$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{49}$algebra. Then $(xx*y)z=x(xy*z)$. With $y=0$, we have $0*z=x*xz$. Put $x=z$ in the last equation to get $0*z=z$ as required.
 This is similar to the proof for ${F}_{10}$algebra.
 Let X be an ${F}_{51}$algebra. Then $yz*xx=(yz*x)x$. With $z=0$, we have $y=(y*x)x$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{53}$algebra. Then $yz*xx=y(zx*x)$ which becomes $yz=y(zx*x)$. Put $z=x$ to get $yx=y*0x$. Substituting $y=x$, we get $x*0x=0$. So, $0x=0\Rightarrow x=0$, which means that X is psemisimple (Theorem 8(2)). Now, put $y=0$ in $yx=y*0x$ to get $0x=0*0x$. Hence, X is quasiassociative. Thus, X is associative.
 Let X be an ${F}_{57}$algebra. Then $(yz*x)x=y(z*xx)$. With $z=0$, we have $yx*x=y$. Put $x=y$ in the last equation to get $0*y=y$ as required.
 Let X be an ${F}_{58}$algebra. Then $(y*zx)x=y(zx*x)$. Put $y=x=z$ to get $x*0x=0$. So, $0x=0\Rightarrow x=0$, which means that X is psemisimple (Theorem 8(2)). Now, put $z=x,y=0$ to get $0x=0*0x$. Hence, X is quasiassociative. Thus, X is associative.
 Let X be an ${F}_{60}$algebra. Then $y(zx*x)=y(z*xx)$. Put $y=x=z$ to get $x*0x=0$. So, $0x=0\Rightarrow x=0$, which means that X is psemisimple (Theorem 8(2)). Hence, X has the LCL by Theorem 4. Thence, the ${F}_{10}$ identity becomes $zx*x=z*xx$. Now, substitute $z=x$ to get $0x=x$. Thus, X is associative.





 
 
 
 

 1.
 Let X be an ${F}_{3}$algebra. X is associative if and only if $x(x*zx)=xz$ if and only if X is psemisimple.
 2.
 Let X be an ${F}_{5}$algebra. X is associative if and only if $(xy*x)x=yx$.
 3.
 Let X be an ${F}_{21}$algebra. X is associative if and only if $(yx*x)x=x*y$.
 4.
 Let X be an ${F}_{42}$algebra. X is associative if and only if X is psemisimple.
 5.
 Let X be an ${F}_{55}$algebra. X is associative if and only if $\left[\right(y*x)*x]*x=x*y$.
 6.
 (a)
 X is an ${F}_{5}$algebra and psemisimple if and only if X is associative.
 (b)
 Let X be an ${F}_{8}$algebra. X is associative if and only if $x(y*zx)=yz$.
 7.
 Let X be an ${F}_{19}$algebra. X is associative if and only if quasiassociative.
 8.
 X is an ${F}_{39}$algebra and obeys $y(x*xz)=zy$ if and only if X is associative.
 9.
 Let X be a ${F}_{46}$algebra. X is associative if and only if $0(0*0x)=x$.
 10.
 (a)
 X is an ${F}_{52}$algebra and ${F}_{55}$algebra if and only if X is associative.
 (b)
 X is an ${F}_{52}$algebra and obeys $(y*zx)x=zy$ if and only if X is associative.
 (c)
 X is an ${F}_{55}$algebra and psemisimple if and only if X is associative.
 (d)
 Let X be an ${F}_{52}$algebra. X is associative if and only if X is quasiassociative.
 11.
 (a)
 X is an ${F}_{59}$algebra and ${F}_{55}$algebra if and only if X is associative.
 (b)
 X is an ${F}_{52}$algebra and obeys $(y*zx)x=zy$ if and only if X is associative.
 (c)
 Let X be a ${F}_{56}$algebra. X is associative if and only if X is quasiassociative.
 (d)
 Let X be an ${F}_{59}$algebra. X is associative if and only if X is quasiassociative.
 Suppose X is a ${F}_{3}$algebra. Then, $xy*zx=x(y*zx)$. Put $y=x$ to get $0*zx=x(x*zx)$. Substituting $x=0$, we have $0z=0*0z$ which means X is quasiassociative. Going by Theorem 9, X is associative if and only if X is psemisimple. Furthermore, by Theorem 4(3) and $0*zx=x(x*zx)$, an ${F}_{3}$algebra X is associative if and only if $xy=x(x*zx)$.
 Suppose X is associative. Then $0*x=x$. X is ${F}_{5}$ implies $(xy*z)x=(x*yz)x$. With $z=x$, we have $(xy*x)x=(x*yx)x\Rightarrow (xy*x)x=(x*x)yx\Rightarrow (xy*x)x=0*yx\Rightarrow (xy*x)x=yx$ as required. Conversely, suppose $(xy*x)x=yx$. Put $z=x$ in $(xy*z)x=(x*yz)x$ to get $(xy*x)x=(x*yx)x\Rightarrow (xy*x)x=(x*x)yx\Rightarrow (xy*x)x=0*yx\Rightarrow yx=0*yx$ (since $(xy*x)x=yx$). So, X is associative.
 Suppose X is associative. Then $x*y=y*x$. X is ${F}_{21}$ implies $yx*zx=(yx*z)x$. With $z=x$, we have $(yx*x)x=y*x=x*y$ as required. Conversely, suppose $(yx*x)x=x*y$. Put $z=x$ in ${F}_{21}$ to get $(yx*x)x=y*x$. So, $x*y=y*x$ as required.
 Suppose X is associative. Then $0*z=z$. X is ${F}_{42}$ implies $xx*yz=(xx*y)z$. With $y=0$, we have $0*0z=0*z=z$ as required. Conversely, suppose $0*0z=z$. Put $y=0$ in ${F}_{42}$ to get $0*0z=0*z$. So, $0*z=z$ as required.
 Suppose X is associative. Then $x*y=y*x$. X is ${F}_{55}$ implies $\left[\right(y*z)*x]*x=[y*(z*x\left)\right]*x$. With $z=x$, we have $\left[\right(y*x)*x]*x=y*x=x*y$ as required. Conversely, suppose $\left[\right(y*x)*x]*x=x*y$. Put $z=x$ in ${F}_{55}$ to get $y*x=\left[\right(y*x)*x]*x=x*y$. So, $y*x=x*y$ as required.
3. Summary, Conclusions and Recommendations
 A loop is an extra loop if and only if the loop is both a Moufang loop and a Cloop.
 A loop is a Moufang loop if and only if the loop is both a right Bol loop and a left Bolloop.
 A loop is a Cloop if and only if the loop is both a RCloop and a LCloop.
 X is an ${F}_{i}$algebra and ${F}_{j}$algebra if and only if X is associative, for the pairs: $i=52,\phantom{\rule{3.33333pt}{0ex}}j=55$, $i=59,\phantom{\rule{3.33333pt}{0ex}}j=55$.
Author Contributions
Funding
Conflicts of Interest
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Fenyves  ${\mathit{F}}_{\mathit{i}}\equiv \mathit{ASS}$  ${\mathit{F}}_{\mathit{i}}\not\equiv \mathit{ASS}$  Quassigroup  ${\mathit{F}}_{\mathit{i}}+\mathit{BCI}$ 

Identity  Inaloop  Inaloop  ⇒ Loop  ⇒ ASS 
${F}_{1}$  ✓  ✓  ✓  
${F}_{2}$  ✓  ✓  ✓  
${F}_{3}$  ✓  ✓  ‡  
${F}_{4}$  ✓  ✓  
${F}_{5}$  ✓  ‡  
${F}_{6}$  ✓  ✓  ✓  
${F}_{7}$  ✓  ✓  
${F}_{8}$  ✓  ‡  
${F}_{9}$  ✓  ✓  
${F}_{10}$  ✓  ✓  
${F}_{11}$  ✓  ✓  ✓  
${F}_{12}$  ✓  ✓  ✓  
${F}_{13}$  ✓  ✓  ✓  
${F}_{14}$  ✓  ✓  
${F}_{15}$  ✓  ✓  
${F}_{16}$  ✓  ✓  
${F}_{17}$  ✓  ✓  ✓  
${F}_{18}$  ✓  ✓  ✓  
${F}_{19}$  ✓  ‡  
${F}_{20}$  ✓  ✓  
${F}_{21}$  ✓  ✓  ‡  
${F}_{22}$  ✓  ✓  ✓  
${F}_{23}$  ✓  ✓  
${F}_{24}$  ✓  ✓  
${F}_{25}$  ✓  ✓  
${F}_{26}$  ✓  ✓  
${F}_{27}$  ✓  ✓  ✓  
${F}_{28}$  ✓  ✓  ✓  
${F}_{29}$  ✓  ‡  
${F}_{30}$  ✓  ✓  
${F}_{31}$  ✓  ✓  ✓  
${F}_{32}$  ✓  ✓  ✓  
${F}_{33}$  ✓  ✓  
${F}_{34}$  ✓  ✓  
${F}_{35}$  ✓  ✓  
${F}_{36}$  ✓  ✓  
${F}_{37}$  ✓  ✓  
${F}_{38}$  ✓  ✓  ✓  
${F}_{39}$  ✓  ‡  
${F}_{40}$  ✓  ✓  
${F}_{41}$  ✓  ✓  ✓  
${F}_{42}$  ✓  ‡  
${F}_{43}$  ✓  ✓  
${F}_{44}$  ✓  ✓  
${F}_{45}$  ✓  ✓  
${F}_{46}$  ✓  ‡  
${F}_{47}$  ✓  ✓  ✓  
${F}_{48}$  ✓  ✓  
${F}_{49}$  ✓  ✓  
${F}_{50}$  ✓  ✓  
${F}_{51}$  ✓  ✓  
${F}_{52}$  ✓  ‡  
${F}_{53}$  ✓  ✓  ✓  
${F}_{54}$  ✓  ‡  
${F}_{55}$  ✓  ‡  
${F}_{56}$  ✓  ‡  
${F}_{57}$  ✓  ✓  
${F}_{58}$  ✓  ✓  ✓  
${F}_{59}$  ✓  ‡  
${F}_{60}$  ✓  ✓ 
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Jaíyéọlá, T.G.; Ilojide, E.; Olatinwo, M.O.; Smarandache, F. On the Classification of BolMoufang Type of Some Varieties of Quasi Neutrosophic Triplet Loop (Fenyves BCIAlgebras). Symmetry 2018, 10, 427. https://doi.org/10.3390/sym10100427
Jaíyéọlá TG, Ilojide E, Olatinwo MO, Smarandache F. On the Classification of BolMoufang Type of Some Varieties of Quasi Neutrosophic Triplet Loop (Fenyves BCIAlgebras). Symmetry. 2018; 10(10):427. https://doi.org/10.3390/sym10100427
Chicago/Turabian StyleJaíyéọlá, Tèmítọ́pẹ́ Gbọ́láhàn, Emmanuel Ilojide, Memudu Olaposi Olatinwo, and Florentin Smarandache. 2018. "On the Classification of BolMoufang Type of Some Varieties of Quasi Neutrosophic Triplet Loop (Fenyves BCIAlgebras)" Symmetry 10, no. 10: 427. https://doi.org/10.3390/sym10100427