# Inverse Properties in Neutrosophic Triplet Loop and Their Application to Cryptography
## 2 Main Results
For a groupoid \(X\): \(L_{x}\) is bijective for all \(x\in X\) if and only if \(X\) is a left quasigroup and \(R_{x}\) is bijective for all \(x\in X\) if and only if \(X\) is a right quasigroup. Hence, (a) to (d) are equivalent. If \(X\) is a WIPNTL, then it is both an LWIPNTL and RWIPNTL which implies that \(L_{x}JR_{x}=J\) and \(R_{x}JL_{x}=J\). Consequently, \(L_{x}JR_{x}^{2}JL_{x}=J^{2}\) and \(R_{x}JL_{x}^{2}JR_{x}=J^{2}\). Thus, \(L_{x}^{2}=I\Leftrightarrow R_{x}^{2}=I\). 5. This follows from Lemma 2. 6. This follows from Lemma 2. Theorem 7: _Let \(X\) be an NTL._  
1. \(X\) _has the LWIP and AAIP, then_ \(X\) _has the RIP._
2. \(X\) _has the RWIP and AAIP, then_ \(X\) _has the LIP._
3. \(X\) _has the LWIP and AIP, then_ \(X\) _has the RCIP._
4. \(X\) _has the RWIP and AIP, then_ \(X\) _has the LCIP._
5. \(X\) _is an IPNTL if and only if_ \(X\) _is a WIPNTL and an AAIPNTL._
6. \(X\) _is a CIPNTL if and only if_ \(X\) _is a WIPNTL and an AIPNTL._  
Proof. : Let \(X\) be an NTL. 1. LWIP implies \(anti(xy)x=anti(y)\stackrel{{\text{AAIP}}}{{\Rightarrow}}anti(y)anti (x)*x=anti(y)\stackrel{{\text{y}\rightarrow\text{anti}(y)}}{{ \Rightarrow}}anti(x)\) \(anti(anti(x))*anti(x)=anti(anti(y))\Rightarrow yx*anti(x)=y\Rightarrow\) RIP. 2. RWIP implies \(x\)\(anti(yx)=anti(y)\stackrel{{\text{AAIP}}}{{\Rightarrow}}x*anti(x)anti (y)=anti(y)\stackrel{{\text{y}\rightarrow\text{anti}(y)}}{{ \Rightarrow}}x\rightarrowanti(x)\) \(anti(x)*anti(x)anti(anti(y))=anti(anti(y))\Rightarrow anti(x)*xy=y\Rightarrow\) LIP. 3. LWIP implies \(anti(xy)x=anti(y)\stackrel{{\text{AIP}}}{{\Rightarrow}}anti(x)anti (y)*x=anti(y)\) \(anti(x)\) \(anti(x)\) \(anti(x)\) \(anti(x)=anti(anti(y))\Rightarrow xy*anti(x)=y\Rightarrow\) RCIP. 4. RWIP implies \(x\)\(anti(yx)=anti(y)\stackrel{{\text{AIP}}}{{\Rightarrow}}x*anti(y)anti (x)=anti(y)\) \(\stackrel{{\text{y}\rightarrow\text{anti}(y)}}{{\Rightarrow}} \)\(anti(x)*anti(x)\) \(anti(x)*anti(x)\) \(anti(x)\) \(anti(x)\) \(anti(x)\) \(anti(y)anti(anti(x))=anti(anti(y))\Rightarrow anti(x)*yx=y\Rightarrow\) LCIP. 5. This backward of the statement follows by 1 and 2, while the forward of the statement follows by 1 of Theorem 5. 6. This backward of the statement follows by 3 and 4, while the forward of the statement follows by 2 of Theorem 5. Lemma 6: _Let \(X\) be an NTL._  
1. _If_ \(X\) _is an AIPNTL, then_ 1. \(neut\big{(}anti(x)\big{)}=anti\big{(}neut(x)\big{)}\)_._ 2. _If_ \(X\) _is an AAIPNTL, then_ 1. \(neut\big{(}anti(x)\big{)}=anti\big{(}neut(x)\big{)}\)_._ 2. _If_ \(X\) _is an AAIPNTL, then_ 1. \(neut\big{(}anti(x)\big{)}=anti\big{(}neut(x)\big{)}\)_._ 2. _until_ 2. _until_ 2. _until_ 2  
**Proof. **  
1. Let \(X\) be an AIPNTL. Then, \(anti(xy)=anti(x)anti(y)\). 1. Put \(y=neut(x)\) to get \(anti\big{(}x\ neut(x)\big{)}=anti(x)anti\big{(}neut(x)\big{)}\Rightarrow\) \[x\)\(anti(neut(x))=x.\]4 (28) Do the replacement \(x\mapsto neut(x)\) and put \(y=x\) to get \(anti\big{(}neut(x)x\big{)}=anti\big{(}neut(x)\big{)}anti(x)\Rightarrow\) \[x\)\(anti(neut(x))=x.\]5 (29) Combining (28) and (29), we get \(neut\big{(}anti(x)\big{)}=anti\big{(}neut(x)\big{)}\). 2. Do the replacements \(x\mapsto neut(x)\) and \(y\mapsto neut(y)\) to get \[x\)\(anti(neut(x))=x.\]6
2. Let \(X\) be an AAIPNTL. Then, \(anti(xy)=anti(y)anti(x)\). 1. Put \(y=neut(x)\) to get \(anti\big{(}x\ neut(x)\big{)}=anti\big{(}neut(x)\big{)}anti(x)\Rightarrow\) \[x\)\(anti(neut(x))=x.\]7 (30) Do the replacement \(x\mapsto neut(x)\) and put \(y=x\) to get \(anti\big{(}neut(x)x\big{)}=anti(x)anti\big{(}neut(x)\big{)}\Rightarrow\) \[x\)\(anti(neut(x))=x.\]8 (31) Combining (30) and (31), we get \(neut\big{(}anti(x)\big{)}=anti\big{(}neut(x)\big{)}\). 2. Do the replacements \(x\mapsto neut(x)\) and \(y\mapsto neut(y)\) to get \[x\)\(anti(neut(x))=x.\]9
3. This follows from the AIP and AAIP. 4. This follows from the AIP and AAIP. **Theorem 8**. : _Let \((\mathbb{Z}_{p},+,\cdot)\) be the field of integers modulo \(p\), where \(p\) is prime. Define \(*\) on \(\mathbb{Z}_{p}\) as follows: \(x*y=ax+ay\) for a fixed \(0,1\
eq a\in\mathbb{Z}_{p}\). Then: 1.