# Section 2.5: Problem 2 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
Prove the equivalence of parts (a) and (b) of the completeness theorem.Suggestion: $\Gamma\vDash\phi$ iff $\Gamma\cup\{\neg\phi\}$ is unsatisfiable. And $\Delta$ is satisfiable iff $\Delta\not\vDash\bot$ , where $\bot$ is some unsatisfiable, refutable formula like $\neg\forall xx=x$ .
Remark: Similarly, the soundness theorem is equivalent to the statement that every satisfiable set of formulas is consistent.
Completeness Theorem. (a) If $\Gamma\vDash\phi$ , then $\Gamma\vdash\phi$ .
(b) If $\Gamma$ is consistent, then $\Gamma$ is satisfiable.
(a) implies (b). If $\Gamma$ is not satisfiable, then $\Gamma\vDash\{\phi,\neg\phi\}$ , hence, $\Gamma\vdash\{\phi,\neg\phi\}$ , and $\Gamma$ is not consistent.
(b) implies (a). If $\Gamma\vDash\phi$ , then $\vDash_{\mathfrak{A}}\Gamma[s]$ implies $\vDash_{\mathfrak{A}}\phi[s]$ and $\not\vDash_{\mathfrak{A}}\neg\phi[s]$ , i.e.$\Gamma;\neg\phi$ is unsatisfiable, and, hence, inconsistent, implying, by RAA, $\Gamma\vdash\phi$ .