Check whether a set of formulas entails a wff

Question

Given Γ = {p, p → q, q → ¬p, ¬(r ↔ q)} and α = r ∨ q.

I have to check whether Γ |= α for the given Γ and α. My solution -

Let V be an arbitrary valuation such that V |= Γ.

This implies V |= p and V |= (p → q).

Therefore, V(p) = T and V(p → q) = T.

Therefore, V(q) = T.

Now, V(p) = T and V(q) = T.

Therefore, V(q → ¬p) = F.

Does this mean that Γ is unsatisfiable, and thus Γ does not entail α? Correct me if I'm wrong.

Answer 1

The argument is fine but the conclusion is wrong.

Due to the fact that the set $\Gamma$ is unsatisfiable (as you have shown), and due to the definition of logical implication (in this case : tautological implication) [see Herbert Enderton, A Mathematical Introduction to Logic (2nd ed, 2001), page 23] :

DEFINITION. $\Gamma$ tautologically implies $\alpha$ (written : $\Gamma \vDash \alpha$) iff every truth assignment for the sentence symbols in $\Gamma$ and $\alpha$ that satisfies every member of $\Gamma$ also satisfies $\alpha$.

we have that in the case in which no truth assignment satisfies every member of $\Gamma$, then for any $\alpha$, it is vacuously true that $\Gamma \vDash \alpha$.

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Note : see Vacuous truth.

We have that the above definition is a conditional :

for every truth assignement $v$, if $v$ satisfy every formula $\gamma \in \Gamma$, then $v$ satisfy $\alpha$.

In our example, the antecedent is false; thus, the conditional is true.