# Inverse Properties in Neutrosophic Triplet Loop and Their Application to Cryptography
## 2 Main Results  
Lemma 1: _Let \(X\) be a CIPNTL. Then:_  
1. \(neut(x)=neut(anti(x))\)_,_ \(anti(anti(x))=x\) _and_ \(J^{2}=I\)_._
2. \(L_{x}R_{anti(x)}=I=R_{x}L_{anti(x)}\)_X is an RIPNTL if and only X is an LIPNTL._
4. \(neut(x)=anti(neut(x))\) _and neut(neut(x)) = neut(x)neut(x)._  
**Proof. **  
1. Put \(y=anti(x)\) in (4) to get \(x\)\(anti(x)*anti(x)=anti(x)\Rightarrow\) \[neut(x)anti(x)=anti(x)\] (10) Put \(y=anti(x)\) in (3) to get \(anti(x)*anti(x)x=anti(x)\Rightarrow\) \[anti(x)neut(x)=anti(x)\] (11) By (10) and (11), we have \(neut(x)=neut(anti(x))\). By this, \(anti(x)x=x\)\(anti(x)=neut(x)\Rightarrow anti(anti(x))=x\) and \(J^{2}=I\). 2. These are just (3) and (4) put in translation forms. 3. From 2., \(L_{x}R_{anti(x)}R_{x}L_{anti(x)}=I\). So, \(R_{anti(x)}R_{x}=I\Rightarrow L_{x}R_{anti(x)}\Leftrightarrow y\)\(anti(x)*x=y\Rightarrow anti(x)*xy=y\Leftrightarrow y\)\(anti(x)*anti(x)=y\Rightarrow anti(x)*xy=y\Leftrightarrow yx*anti(x)=y\Rightarrow anti(x)*xy=y\Leftrightarrow X\) has the RIP, which implies that X has the LIP. Similarly, since by 2., \(R_{x}L_{anti(x)}L_{x}R_{anti(x)}=I\), then we get X has the LIP implies X has the RIP. 4. Let \(x\in X\). Recall that \(x\)\(neut(x)=x=neut(x)x\). So, by the RCIP, \(neut(x)x*anti(neut(x))=x\)\(anti(neut(x))\Rightarrow\) \[x\)\(anti(neut(x))=x.\] (12) Similarly, by the LCIP, \[anti(neut(x))x=x.\] (13) Thus, by (12) and (13), \(neut(x)=anti(neut(x))\). Furthermore, \(neut(x)neut(x)=anti(neut(x))neut(x)=neut(x)anti(neut(x))\Rightarrow neut(neut(x ))=neut(x)neut(x)\). \(\square\)  
**Lemma 2**. : _Let X be a CIPNTL or an IPLNTL. Then:_  
1. _Equations_ \(a*x=b\) _and_ \(y*c=d\) _have solutions for_ \(x,y\in X\) _and these solutions are unique for all_ \(a,b,c,d\in X\)_. (unique solvability)_
2. _The cancellation laws hold._
3. _The right and left translation maps_ \(R_{a}\) _and_ \(L_{a}\) _are bijections for all_ \(a\in X\)_._  
**Proof. ** For CIPNTL. 1. \(a*x=b\Rightarrow(a*x)anti(a)=b\)\(anti(a)\Rightarrow x=b\)\(anti(a)\in X.\) Similarly, \(y*c=d\Rightarrow anti(c)(y*c)=anti(c)d\Rightarrow y=anti(c)d\). Let \(x_{1},x_{2}\in X\) such that \(a*x_{1}=b=a*x_{2}\Rightarrow(a*x_{1})anti(a)=(a*x_{2})anti(a)\Rightarrow x_{1 }=x_{2}.\)
2. This follows from 1. 3. \(R_{a}\ :\ X\to X\) given by \(xR_{a}=x*a\). \(R_{a}\) is a bijection if and only if the equation \(x*a=b\) is uniquely solvable for \(x\) for all \(a,b\in X\). \(L_{a}\ :\ X\to X\) given by \(xL_{a}=a*x\). \(L_{a}\) is a bijection if and only if the equation \(a*x=b\) is uniquely solvable for \(x\) for all \(a,b\in X\). For IPNTL. 1. \(a*x=b\Rightarrow anti(a)(a*x)=anti(a)b\Rightarrow x=anti(a)b\in X.\) Similarly, \(y*c=d\Rightarrow(y*c)anti(c)=d\)\(anti(c)\Rightarrow y=d\)\(anti(c)\). Let \(x_{1},x_{2}\in X\) such that \(a*x_{1}=b=a*x_{2}\Rightarrow anti(a)(a*x_{1})=anti(a)(a*x_{2})\Rightarrow x_{1 }=x_{2}.\)This follows from above. 3. \(R_{a}\ :\ X\to X\) given by \(xR_{a}=x*a\). \(R_{a}\) is a bijection if and only if the equation \(x*a=b\) is uniquely solvable for \(x\) for all \(a,b\in X\). \(L_{a}\ :\ X\to X\) given by \(xL_{a}=a*x\). \(L_{a}\) is a bijection if and only if the equation \(a*x=b\) is uniquely solvable for \(x\) for all \(a,b\in X\). **Theorem 2**. : _Let \(X\) be an NTL._  
1. \(X\) _is an_ _RCIPNTL_ _if and only if_ \(x*y\ anti(x)=y\)_._
2. \(X\) _is an_ _LCIPNTL_ _if and only if_ \(anti(x)y*x=y\)_._
3. \(X\) _is a_ _CIPNTL_ _if and only if_ \(x*y\ anti(x)=y=anti(x)y*x\)_._  
Proof. :
1. By Lemma 2, if \(X\) is an RCIPNTL, then it is a left quasigroup and \(L_{a}\) is a bijection for \(a\in X\). Consider an NTL which has the property \(x*y\ anti(x)=y\). Put \(y=neut(anti(x))\) to get \(x*neut(anti(x))anti(x)=neut(anti(x))\Rightarrow x*anti(x)=neut(anti(x)) \Rightarrow neut(x)=neut(anti(x))\Rightarrow anti(anti(x))=x\). Thus, \(x*a=b\Rightarrow x*anti(anti(a))=b\Rightarrow anti(a)(x*anti(ati(a)))=anti(a)b \Rightarrow x=anti(a)b\). Let \(x_{1},x_{2}\in X\).