A query regarding P vs NP

Question

Problem Statement: Given a Problem $(A)$ such that: Any proposed solution $(S_0)$ is guaranteed to be polynomial size w.r.t. $A$. Now, consider a fixed set of $K$ properties $\{k_0, k_1,...k_c\}$, where each property is $coNP-Complete$.

For any solution $S_0$ is considered a valid solution iff: "None of the property in $K$ has a $NO$ instance for $S_0$".

Example: For some hypothetical problem $(A')$ let the solution $S_0'$ be a simple graph with $n$ nodes. Let $k_i$ be the property: "$S_0'$ does not contain a clique with $n/4$ nodes".

Query: Assuming the highly improbable scenario that $P=NP$, it implies $P=NP=co-NP)$. Thus, all the properties in $K$ are $P-Complete$. Does that imply that finding the solution $(S_0)$ for $(A)$ is in $P$ ? It seems so but still need confirmation.

Answer 1

Your problem is essentially a $NP^{coNP}$ problem. If $P=NP$ then the polynomial hierarchy collapses and your problem is in $P$.

If $P=NP=coNP$ then we can check in polynomial time whether a solution $S_0$ is valid (using a polynomial-time algorithm to check each property, which exists because $coNP=P$). Thus, the problem is in $NP$, and thus, in $P$.