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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 .$P^{SAT}⊆NP$As$coNP⊆P^{SAT}$is already known, which result in$coNP⊆NP$ $coNP⊆NP$

• 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:

$L∈P^{SAT}→L∈P^{SAT}$

• 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.$L∈NP→\stackrel{¯}{L}∈NP$ , that really is exactly the same except$L∈NP→L∈coNP$ from the definition of.$coNP$

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