# Inverse Properties in Neutrosophic Triplet Loop and Their Application to Cryptography
## 2 Main Results
\(X\) _is normal if and only if_ \((x,y)anti(xy)*(yx)=neut(y)neut(x)\) _if and only if_ \((yx)*(x,y)anti(xy)=neut(y)neut(x)\) _if and only if_ \((yx)*anti(xy)[x,y]=neut(y)neut(x)\)_._
14. \(\big{(}x,neut(x),anti(x)\big{)}=neut(neut(x))=\big{(}anti(x),neut(x),x\big{)}\)_._
15. \(\big{(}neut(x),x,anti(x)\big{)}=\big{(}anti(x),x,neut(x)\big{)}=\big{(}x,anti(x),neut(x)\big{)}=\big{(}neut(x),anti(x),x\big{)}=neut(x)\)_._  
Proof. :
**1 and 2**: From the right associator, \(xy*anti(x)=\big{(}x*y\ anti(x)\big{)}\big{(}x,y,anti(x)\big{)}\Rightarrow y=y \big{(}x,y,anti(x)\big{)}\Rightarrow y\ anti(y)=y\big{(}x,y,anti(x)\big{)}*anti(y) \Rightarrow\big{(}x,y,anti(x)\big{)}=neut(y)\). Hence, \(\big{(}x,x,anti(x)\big{)}=neut(x)\). From the left associator, \(xy*anti(x)=\big{[}x,y,anti(x)\big{]}\big{(}x*y\ anti(x)\big{)}\Rightarrow y= \big{[}x,y,anti(x)\big{]}y\Rightarrow anti(y)*\big{[}x,y,anti(x)\big{]}y \Rightarrow[x,y,anti(x)]=neut(y)\). Hence, \(\big{[}x,x,anti(x)\big{]}=neut(x)\). **3 and 4**: From the right associator, \(anti(x)y*x=\big{(}anti(x)*yx\big{)}\big{(}anti(x),y,x)\Rightarrow y=y\big{(}anti(x),y,x\big{)}\Rightarrow y\ anti(y)=y\big{(}anti(x),y,x\big{)}*anti(y)\Rightarrow \big{(}anti(x),y,x\big{)}=neut(y)\). Hence, \(\big{(}anti(x),x,x\big{)}=neut(x)\). From the left associator, \(anti(x)y*x=\big{[}anti(x),y,x\big{]}\big{(}anti(x)*yx\big{)}\Rightarrow y=\big{[} anti(x),y,x\big{]}y\Rightarrow\)\(anti(y)y=anti(y)*[anti(x),y,x]y\Rightarrow[anti(x),y,x]=neut(y)\). Hence, \([anti(x),x,x]=neut(x)\). 5. From the right commutator, \(x*neut(x)=(neut(x)*x)(x,neut(x))\Rightarrow x=x(x,neut(x))\Rightarrow x\)\(anti(x)=x(x,neut(x))*anti(x)\Rightarrow(x,neut(x))=neut(x)\). Similarly, \((neut(x),x)=neut(x)\). From the left commutator, \(x*neut(x)=[x,neut(x)](neut(x)*x)\Rightarrow x=[x,neut(x)]x\Rightarrow anti(x)x= anti(x)x=anti(x)*x(x,neut(x))\Rightarrow[x,neut(x)]=neut(x)\). Similarly, \([neut(x),x]=neut(x)\). 6. From the right commutator, \(x*x=(xx)(x,x)\Rightarrow xx*anti(xx)=(xx)(x,x)*anti(xx)\Rightarrow neut(xx)=(x,x)\). From the left commutator, \(x*x=[x,x](xx)\Rightarrow anti(xx)*xx=anti(xx)*[x,x](xx)\Rightarrow neut(xx)=[x,x]\). 7. From the right commutator, \(x*anti(x)=(anti(x)*x)(x,anti(x))\Rightarrow neut(x)(x,anti(x))\Rightarrow neut(x) (x,anti(x))\Rightarrow neut(x)(x,anti(x))\Rightarrow big(x,anti(x))=neut( neut(x))\). Similarly, \((anti(x),x)=neut(neut(x))\). From the left commutator, \(anti(x)*x=[x,anti(x)](x*anti(x))\Rightarrow neut(x)=[x,anti(x)]neut(x)\Rightarrow anti (neut(x))neut(x)=anti(neut(x))*[x,anti(x)]neut(x)\Rightarrow[x,anti(x)]=neut( neut(x))\). Similarly, \([anti(x),x]=neut(neut(x))\). 8. From the right commutator, \(xy=yx*(x,y)\Rightarrow xy*anti(yx)=(yx)(x,y)*anti(yx)\Rightarrow(x,y)=(xy)anti(yx)\). From the left commutator, \(xy=[x,y]*yx\Rightarrow anti(yx)*xy=anti(yx)*[x,y](yx)\Rightarrow[x,y]=anti(yx)(xy)\). 9. This follows from 6 and 10. 10. We shall use 10. \(X\) is normal if and only if \((x,y)anti(xy)=(xy)anti(yx)*anti(yx)\Leftrightarrow(x,y)anti(xy)=anti(yx)\Leftrightarrow\)  
\((x,y)anti(xy)*(yx)=anti(yx)(yx)\) or \((yx)*(x,y)anti(xy)=(yx)anti(yx)\Leftrightarrow\)  
\((x,y)anti(xy)*(yx)=neut(yx)\) or \((yx)*(x,y)anti(xy)=neut(yx)\Leftrightarrow\)  
\((x,y)anti(xy)*(yx)=neut(y)neut(x)\) or \((yx)*(x,y)anti(xy)=neut(y)neut(x)\). \(X\) is normal if and only if \(anti(xy)[x,y]=anti(xy)*anti(yx)(xy)\Leftrightarrow anti(xy)[x,y]=anti(yx)\Leftrightarrow\)  
\(anti(xy)[x,y]*(yx)=anti(yx)(yx)\) or \((yx)*anti(xy)[x,y]=(yx)anti(yx)\Leftrightarrow\)  
\(anti(xy)[x,y]*(yx)=neut(yx)\) or \((yx)*anti(xy)[x,y]=neut(yx)\Leftrightarrow\)  
\(anti(xy)[x,y]*(yx)=neut(y)neut(x)\) or \((yx)*anti(xy)[x,y]=neut(y)neut(x)\). 14. Apply the right and left associators. 15. Apply the right and left associators. Lemma 4: _Let \(X\) be an NTL._  
1. _Let_ \(X\) _be an RIPNL._ \(X\) _is a CIPNTL if and only if_ \(X\) _is commutative._
2. _Let_ \(X\) _be an LIPNL._ \(X\) _is a CIPNTL if and only if_ \(X\) _is commutative._
3. _Let_ \(X\) _be commutative. The following are equivalent:_ 1. _RIP._ 2. _LIP._ 3. _RCIP._ 4.