To show that Γ entails ¬A, we **assume for contradiction** that there were some model Μ of Γ which does not satisfy ¬A. It follows that M satisfies A.

By our assumption in the problem statement, we have that A entails B; this means that **any model which satisfies A will also satisfy B.** Thus, **our model M also satisfies B**.

However, we also assumed (in the problem statement) that Γ entailsc ¬B; since **M is a model of **Γ, **it satisfies ¬B. **This contradicts the line above, so there must be no such model M which satisfies Γ and does not satisfy ¬A. Thus, every model of Γ satisfies ¬A, so Γ entails ¬A.