# Inverse Properties in Neutrosophic Triplet Loop and Their Application to Cryptography
## 2 Main Results
Then, \(x_{1}*a=x_{2}*a\Rightarrow x_{1}*anti(anti(a))=x_{2}*anti(anti(a))\Rightarrow anti (a)\big{(}x_{1}*anti(anti(a))\big{)}=anti(a)\big{(}x_{2}*anti(anti(a))\big{)} \Rightarrow x_{1}=x_{2}\). So, \(x*a=b\) is uniquely solvable for \(x\) that \(R_{a}\) is bijective. RCIP implies \(L_{x}R_{anti(x)}=I\Rightarrow R_{anti(x)}=L_{x}^{-1}\Rightarrow R_{anti(x)}L_{ x}=I\Rightarrow x*y\ anti(x)=y\). Conversely, \(x*y\ anti(x)=y\Rightarrow R_{anti(x)}L_{x}=I\Rightarrow L_{x}=R_{anti(x)}^{-1} \Rightarrow L_{x}R_{anti(x)}=I\Rightarrow\) RCIP. 2. By Lemma 2, if \(X\) is an LCIPNTL, then it is a right quasigroup and \(R_{a}\) is a bijection for \(a\in X\). Consider an NTL which has the property \(anti(x)y*x=y\). Put \(y=neut(anti(x))\) to get \(anti(x)neut(anti(x))*x=neut(anti(x))\Rightarrow anti(x)*x=neut(anti(x)) \Rightarrow neut(x)=neut(anti(x))\Rightarrow anti(anti(x))=x\). Thus, \(a*x=b\Rightarrow anti(anti(a))*x=b\Rightarrow(anti(a))*x)anti(a)=b\ anti(a) \Rightarrow x=b\ anti(a)\). Let \(x_{1},x_{2}\in X\). Then, \(a*x_{1}=a*x_{2}*\Rightarrow anti(anti(a))*x_{1}*anti(a)=anti(anti(a))*x_{2} \Rightarrow\big{(}anti(anti(a))*x_{1}\big{)}anti(a)=\big{(}anti(anti(a))*x_{2} \big{)}anti(a)\Rightarrow x_{1}=x_{2}\). So, \(a*x=b\) is uniquely solvable for \(x\) that \(L_{a}\) is bijective. LCIP implies \(R_{x}L_{anti(x)}=I\Rightarrow R_{x}^{-1}=L_{anti(x)}\Rightarrow L_{anti(x)}R_{ x}=I\Rightarrow anti(x)y*x=y\). Conversely, \(anti(x)y*x=y\Rightarrow L_{anti(x)}R_{x}=I\Rightarrow L_{anti(x)}^{-1}=R_{x} \Rightarrow R_{x}L_{anti(x)}=I\Rightarrow\) LCIP. 3. This follows from 1. and 2. **Lemma 3**. : _Let \(X\) be an IPNL. Then:_  
1. \(neut(x)=neut(anti(x))\)_,_ \(anti(anti(x))=x\) _and_ \(j^{2}=1\)_._
2. \(R_{x}R_{anti(x)}=I=L_{x}L_{anti(x)}\)_._
3. \(X\) _is an_ _RCIPNL_ _if and only_ \(X\) _is an_ _LCIPNL_._
4. \(neut(x)=anti(neut(x))\) _and_ \(neut(neut(x))=neut(x)neut(x)\)_._  
Proof. :
1. Put \(y=anti(x)\) in (1) to get \(anti(x)x*anti(x)=anti(x)\Rightarrow\) \[neut(x)anti(x)=anti(x)\] (14) Put \(y=anti(x)\) in (2) to get \(anti(x)*x\ anti(x)=anti(x)\Rightarrow\) \[anti(x)neut(x)=anti(x)\] (15)By (14) and (15), 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 (1) and (2) put in translation forms. 3. Keep Theorem 2 in mind. From 2., \(R_{x}R_{anti(x)}L_{x}L_{anti(x)}=I\). So, \(R_{anti(x)}L_{x}=I\Rightarrow R_{x}L_{anti(x)}\Leftrightarrow x*y\ anti(x)=y \Rightarrow anti(x)*yx=y\Leftrightarrow\ X\) has the RCIP implies \(X\) has the LCIP. Similarly, since by 2., \(L_{x}L_{anti(x)}R_{x}R_{anti(x)}=I\), then we get \(X\) has the LCIP implies \(X\) has the RCIP. 4. Let \(x\in X\). Recall that \(x\ neut(x)=x=neut(x)\). So, by the RIP, \(x\ neut(x)*anti(neut(x))=x\ anti(neut(x))\Rightarrow\)  
\[x\ anti(neut(x))=x. \tag{16}\]  
Similarly, by the LIP,  
\[anti(neut(x))x=x. \tag{17}\]  
Thus, by (16) and (17), \(neut(x)=anti(neut(x))\). Furthermore, \(neut(x)neut(x)=anti(neut(x))neut(x)=neut(x)anti(neut(x))\Rightarrow neut( neut(x))=neut(x)neut(x)\). Theorem 3.: _Let \(X\) be a CIPNTL. For all \(x,y\in X\),_  
1. \(\big{(}x,x,anti(x)\big{)}=neut(x)=\big{[}x,x,anti(x)\big{]}\)_._
2. \(\big{(}x,y,anti(x)\big{)}=neut(y)=\big{[}x,y,anti(x)\big{]}\)_._
3. \(\big{(}anti(x),x,x\big{)}=neut(x)=\big{[}anti(x),x,x\big{]}\)_._
4. \(\big{(}anti(x),y,x\big{)}=neut(y)=\big{[}anti(x),y,x\big{]}\)_._
5. \(\big{(}x,neut(x)\big{)}=neut(x)=\big{[}x,neut(x)\big{]}\)_._
6. \(\big{(}neut(x),x\big{)}=neut(x)=\big{[}neut(x),x\big{]}\)_._
7. \(\big{(}x,x\big{)}=neut(xx)=[x,x]\)_._
8. \(\big{(}x,anti(x)\big{)}=neut(neut(x))=\big{[}x,anti(x)\big{]}\)_._
9. \(\big{(}anti(x),x\big{)}=neut(neut(x))=\big{[}anti(x),x\big{]}\)_._
10. \((x,y)=(xy)anti(yx)\) _and_ \([x,y]=anti(yx)(xy)\)_._
11. \(X\) _is commutative if and only if_ \((x,y)=neut(yx)\) _if and only if_ \([x,y]=neut(yx)\)_._
12. _If_ \(X\) _is commutative, then_ \(X\) _is normal if and only if_ \((x,y)=(x,neut(x))(y,neut(y))\) _if and only if_ \([x,y]=[x,neut(x)][y,neut(y)]\)_._
13.