Show thatif,.$\mathit{N}\mathit{P}\mathbf{=}{\mathit{P}}^{\mathbf{S}\mathbf{A}\mathbf{T}}$then$NP=coNP$

If $NP=P^{SAT}$that's really the case, all that's left to do now is prove it$NP=coNP$. It's also possible to reduce it straight from either the hypothesis$NP=P^{SAT}$ , that .$P^{SAT}\subseteq NP$As$coNP\subseteq P^{SAT}$is already known, which result in$coNP\subseteq NP$ $coNP\subseteq NP$

• It has always been widely accepted that has been closed as a result of the supplementing operation.,$P^{SAT}$ Because of refuse and approve statuses may simply be swapped. It is possible to deduce that:

$L\in P^{SAT}\to L\in P^{SAT}$

• That predictions can also be used to handle the supplied statement$NP=P^{SAT}$. , even though if that were the scenario, each and every language in the world would be a foreign languagedata-custom-editor="chemistry" $P^{SAT}$.would be in $NP$and vice versa.

• used for which Underneath the supplement operations, it also must be closed.$L\in NP\to \overline{L}\in NP$ , that really is exactly the same except$L\in NP\to L\in coNP$ from the definition of.$coNP$

To look at it another way, based on the information of previous reasoning, it may be concluded that $NP\subseteq coNP$. Hence.$NP=coNP$$NP=coNP$