# New Results on Neutrosophic Extended Triplet Groups Equipped with a Partial Order
## 3 Partially Ordered NETGs
: _(1) If \((G,*,\leq)\) is a po-NETG, then \(P_{G}\subseteq P_{G}^{2}\)._  
_(2) If \((G,*,\leq)\) is a po-WCNETG, then \(P_{G}=P_{G}^{2}\)._  
Proof. : (1) If \((G,*,\leq)\) is a po-NETG, then \(\forall\;a\in P_{G}\), by \(neut(a)\in P_{G}\), we have \(a=a*neut(a)\in P_{G}^{2}\), and so \(P_{G}\subseteq P_{G}^{2}\). (2) If \((G,*,\leq)\) is a po-WCNETG, then \(\forall\;a\in P_{G},\;\forall b\in P_{G}\), by Propositions 3 and 4, we have \(neut(a*b)=neut(b)*neut(a)=neut(a)=neut(a)*neut(b)\leq a*b\), and so \(a*b\in P_{G}\), thus \(P_{G}^{2}\subseteq P_{G}\). Consequently, \(P_{G}=P_{G}^{2}\). **Proposition 8**. : _Let \((G,*,\leq)\) be a po-WCNETG. Then \(\forall\;a\in G,\;aP_{G}a^{-1}\subseteq P_{G}\)._  
Proof. : Let \(a\in G\) and \(b\in P_{G}\), then by Propositions 3 and 4, we have \(neut(a*b*a^{-1})=neut(a^{-1})*neut(a*b)=neut(a*b)*neut(a^{-1})=[neut(b)*neut(a)]*neut (a^{-1})=neut(b)*[neut(a)*neut(a^{-1})]=neut(b)*neut(a^{-1}*a)=neut(b)*neut (neut(a))=neut(b)*(a*a^{-1})=[neut(b)*a]*a^{-1}=[a*neut(b)]*a^{-1}\leq a*b*a^{-1}\), thus \(aba^{-1}\in P_{G}\). Therefore, \(aP_{G}a^{-1}\subseteq P_{G}\). **Lemma 2**. : _Let \((G,*)\) be a WCNETG. Then \(\forall\;a\in G,\;\forall b\in G,\;(a*b)^{-1}=b^{-1}*a^{-1}\)._  
**Proof. ** We know \(a*b\) is an element of \(G\)\(\forall\)\(a\in G\), \(\forall b\in G\) and by Proposition 4, we have \(anti(b)*anti(a)\in\{anti(a*b)\}\). Then using Propositions 1, 5 and Note 1 we get the following identities:  
\[neut(a)=c,\ \{anti(a)\}=\{b\},\ a^{-1}=b;\]8 \[neut(a)=c,\ \{anti(a)\}=\{b\},\ a^{-1}=b;\]9 (Because the multiplication \(*\) is associative) \[neut(b)=c,\ \{anti(b)\}=\{a\},\ b^{-1}=a;\]0 (Because \(G\) is a WCNETG) \[neut(b)=c,\ \{anti(b)\}=\{a\},\ b^{-1}=a;\]1 (Because the multiplication \(*\) is associative) \[neut(b)=c,\ \{anti(b)\}=\{a\},\ b^{-1}=a;\]2 (By Proposition 3) \[neut(b)=c,\ \{anti(b)\}=\{a\},\ b^{-1}=a;\]3  
**Lemma 3**. : _Let \((G,*,\leq)\) be a po-NETG. Then \(P_{G}=P_{N}^{-1}\) and \(P_{G}^{-1}=P_{N}\)._  
**Proof. ** Let \(a\in G\). If \(a\in P_{G}\), then \(neut(a)\leq a\), it follows by Lemma 1 that \(a^{-1}=neut(a^{-1})*a^{-1}=neut(a)*a^{-1}\leq a*a^{-1}=neut(a)=neut(a^{-1})\), and so \(a^{-1}\in P_{N}\), whence \(a=(a^{-1})^{-1}\in P_{N}^{-1}\). Hence, \(P_{G}\subseteq P_{N}^{-1}\). Similarly, we can prove that if \(a\in P_{N}\) then \(a^{-1}\in P_{G}\), so \(P_{N}^{-1}\subseteq P_{G}\). Consequently, \(P_{G}=P_{N}^{-1}\). Similarly, \(P_{G}^{-1}=P_{N}\). **Definition 7**. : _Let \((G,*)\) be a WCNETG. If \(\forall a\in G\), \(\forall b\in G\), \(\forall c\in G\), \(a*neut(c)=b*neut(c)\) implies \(a=b\), then we say \(G\) satisfies neutrosophic cancellation law._  
**Lemma 4**. : _Let \((G,*)\) be a WCNETG satisfying neutrosophic cancellation law and \(P\subseteq G\) satisfy \(\forall\)\(a\in P\), \(a*a=a\). Then \(\forall\)\(a\in G\), \(\forall\)\(b\in G\), \(a*neut(b)\in P\) implies \(neut(a)=a=a^{-1}\)._  
**Proof. ** If \(a*neut(b)\in P\), then \(a*neut(b)=(a*neut(b))*(a*neut(b))=(a*a)*neut(b)\), and so \(a*a=a\), whence \(neut(a)=a\)\(\forall\)\(a\in G\), \(\forall\)\(b\in G\). Then by Lemma 1, we get \(a^{-1}=[neut(a)]^{-1}=neut(a)=a\). **Proposition 9**. : _Let \((G,*)\) be a WCNETG satisfying neutrosophic cancellation law and \(P\subseteq G\) satisfy the following conditions:_  
_(1)_ \(P^{2}\subseteq P\)_:_  
_(2)_ \(P\cap P^{-1}=\{a\in G:neut(a)=a=a^{-1}\}\)_;_  
_(3)_ \(\forall\)\(a\in P\)_,_ \(a*a=a\)_;_  
_(4)_ \(\forall\)\(a\in G\)_,_ \(aPa^{-1}\subseteq P\)_,_  
_then a compatible partial order on \(G\) exists such that \(P\) is the positive cone of \(G\) relative to it. thus  
\[neut(b)=c,\ \{anti(b)\}=\{a\},\ b^{-1}=a;\]4  
However, by Proposition 3, we have  
\[neut(b)=c,\ \{anti(b)\}=\{a\},\ b^{-1}=a;\]5  
and similarly,  
\[neut(b)=c,\ \{anti(b)\}=\{a\},\ b^{-1}=a;\]6  
therefore,  
\[neut(b)=c,\ \{anti(b)\}=\{a\},\ b^{-1}=a;\]7  
and by neutrosophic cancellation law, consequently \(a=b\). Hence, \(\leq\) is anti-symmetric. To prove that \(\leq\) is transitive, let \(a\leq b\) and \(b\leq c\).