# Does the following down-voted answer not answer the question "Why does Schaefer's theorem not prove that P=NP?"?

Does the following highly down-voted answer not answer the question "Why does Schaefer's theorem not prove that P=NP?"? If not, why not?

Marek, V. Wiktor. Introduction to Mathematics of Satisfiability. Boca Raton; London; New-York: CRC Press, 2009, p. 310:

THEOREM 13.3 (Schaefer Theorem)
Let $$Γ$$ be a collection of Boolean tables. If one of conditions (1)–(6) below holds, then $$CSP(Γ)$$ is solvable in polynomial time. Otherwise, it is NP-complete.

If P = NP-complete, the aforementioned conditions have no bearing on the runtime:

If the conditions hold, then $$CSP(Γ)$$ is solvable in P = NP-complete time.
Otherwise, $$CSP(Γ)$$ is solvable in P = NP-complete time.

Which amounts to: "$$CSP(Γ)$$ is solvable in P = NP-complete time."

Review of Introduction to Mathematics of Satisfiability by M. I. Dekhtyar:

Chapter 13 ends with a proof of the well-known Shaefer dichotomy theorem which divides classes of Boolean constraints into solvable in polynomial time and NP-complete.