# Inverse Properties in Neutrosophic Triplet Loop and Their Application to Cryptography
## 2 Main Results
Put \(y=anti(x)\) to get \(anti\big{(}x\;anti(x)\big{)}x=anti\big{(}anti(x)\big{)}\Rightarrow\) \[anti\big{(}neut(x)\big{)}x=anti\big{(}anti(x)\big{)}\] (18) Put \(y=neut(x)\) to get \(anti\big{(}x\;neut(x)\big{)}x=anti\big{(}neut(x)\big{)}\Rightarrow x\;anti(x)= anti\big{(}neut(x)\big{)}\Rightarrow\) \[neut(x)=anti\big{(}neut(x)\big{)}\] (19) (19) implies \(neut(x)neut(x)=anti\big{(}neut(x)\big{)}neut(x)\Rightarrow\) \[anti(x)neut(x)=anti(x)\]0 (20) From (18) and (19), \(neut(x)x=anti\big{(}anti(x)\big{)}\Rightarrow\framebox{\[anti(x)neut(x)=anti(x)\]1 and so, \[anti(x)neut(x)=anti(x)\]2}\) Put \(x=neut(y)\) to get \(anti\big{(}neut(y)\;y)neut(y)=anti(y)\Rightarrow\) \[anti(x)neut(x)=anti(x)\]3 (21)Put \(x=anti(y)\) to get \(anti(anti(y)\ y)anti(y)=anti(y)\Rightarrow anti(neut(y))anti(y)=anti(y)\Rightarrow\)  
\[anti(x)neut(x)=anti(x)\]4  
By (21) and (22), \(\framebox{\[anti(x)neut(x)=anti(x)\]5}\)  
Let \(J:\ X\to X\ \top\ xJ=anti(x)\). Then, \(x_{1}J=x_{2}J\Rightarrow anti(x_{1})=anti(x_{2})\Rightarrow anti(anti(x_{1}))=anti(anti(x_{2}))\Rightarrow x _{1}=x_{2}\). So, \(J\) is 1-1. For all \(y\in X\), there exists \(x\in X\) such that \(xJ=y\) because \(anti(x)=y\Rightarrow anti(anti(x))=anti(y)\Rightarrow x=anti(y)\in X\). Consider \(L_{a}:\ X\to X\ \top\ xL_{a}=ax\). Let \(x_{1}L_{a}=x_{2}L_{a}\Rightarrow ax_{1}=ax_{2}\Rightarrow anti(ax_{1})=anti( ax_{2})\Rightarrow anti(ax_{1})*a=anti(ax_{2})*a\Rightarrow anti(x_{1})=anti(x_{2})\Rightarrow anti(anti(x_{1}))=anti(anti(x_{2}))\Rightarrow x _{1}=x_{2}\). For all \(y\in X\), there exists \(x\in X\) such that \(xL_{a}=y\) because \(ax=y\Rightarrow anti(ax)=anti(y)\Rightarrow anti(ax)*a=anti(y)*a\Rightarrow anti (x)=anti(y)\ a\Rightarrow anti(anti(x))=anti(anti(y)\ a)\Rightarrow x= anti(anti(y)\ a)\). 2. Let \(X\) be an RWIPNTL, then \(x\ anti(yx)=anti(y)\). Put \(y=anti(x)\) to get \(x\ anti(anti(x)x)=anti(anti(x))\Rightarrow\)  
\[anti(x)neut(x)=anti(x)\]6  
Put \(y=neut(x)\) to get \(x\ anti(neut(x)x)=anti(neut(x))\Rightarrow x\ anti(x)=anti(neut(x))\Rightarrow\)  
\[anti(x)neut(x)=anti(x)\]7  
(24) implies \(neut(x)neut(x)=anti(neut(x))neut(x)\Rightarrow\)  
\[anti(x)neut(x)=anti(x)\]8  
From (23) and (24), \(xneut(x)=anti(anti(x))\Rightarrow\framebox{\[anti(x)neut(x)=anti(x)\]9 and so, \[x\)\(anti(neut(x))=x.\]0}\)  
Put \(x=neut(y)\) to get \(neut(y)anti(y\ neu(y))=anti(y)\Rightarrow\)  
\[x\)\(anti(neut(x))=x.\]1  
Put \(x=anti(y)\) to get \(anti(y)anti(y\ anti(y))=anti(y)\Rightarrow anti(y)anti(neut(y))=anti(y)\Rightarrow\)  
\[x\)\(anti(neut(x))=x.\]2  
By (26) and (27), \(\framebox{\[x\)\(anti(neut(x))=x.\]3}\)  
Let \(J:\ X\to X\ \top\ xJ=anti(x)\). Then, \(x_{1}J=x_{2}J\Rightarrow anti(x_{1})=anti(x_{2})\Rightarrow anti(anti(x_{1}))=anti (anti(x_{2}))\Rightarrow x_{1}=x_{2}\). So, \(J\) is 1-1. For all \(y\in X\), there exists \(x\in X\) such that \(xJ=y\) because \(anti(x)=y\Rightarrow anti(anti(x))=anti(y)\Rightarrow x=anti(y)\in X\). Consider \(R_{a}:\ X\to X\ \top\ xR_{a}=xa\). Let \(x_{1}R_{a}=x_{2}R_{a}\Rightarrow x_{1}a=x_{2}a\Rightarrow anti(x_{1}a)=anti(x_{2 }a)\Rightarrow a*anti(x_{1}a)=a*anti(x_{2}a)\Rightarrow anti(x_{1})=anti(x_{2}) \Rightarrow anti(anti(x_{1}))=anti(anti(x_{2}))\Rightarrow x_{1}=x_{2}\). For all \(y\in X\), there exists \(x\in X\) such that \(xR_{a}=y\) because \(xa=y\Rightarrow anti(xa)=anti(y)\Rightarrow a*anti(xa)=a*anti(y)\Rightarrow anti(x)=a \ anti(y)\Rightarrow anti(anti(x))=anti(a\ anti(y))\Rightarrow x=anti(a\ anti(y))\). 3.\(X\) is an LWIPNTL if and only if \(anti(xy)x=anti(y)\Leftrightarrow L_{x}JR_{x}=J\) and \(X\) is an RWIPNTL if and only if \(x\ anti(yx)=anti(y)\Leftrightarrow R_{x}JL_{x}=J\). \(X\) is an LWIPNTL and \(R_{x}\) is bijective if and only if \((L_{x}JR_{x})^{-1}=J^{-1}\) and \(R_{x}\) is bijective if and only if \(R_{x}^{-1}J^{-1}L_{x}^{-1}=J^{-1}\Leftrightarrow R_{x}^{-1}JL_{x}^{-1}=J \Leftrightarrow R_{x}JL_{x}=J\) and \(L_{x}\) is bijective if and only if \(X\) is an RWIPNTL and \(L_{x}\) is bijective.