Metamath Proof Explorer


Theorem pm2.61ne

Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 24-May-2006) (Proof shortened by Andrew Salmon, 25-May-2011) (Proof shortened by Wolf Lammen, 25-Nov-2019)

Ref Expression
Hypotheses pm2.61ne.1 ⊢ A = B → ψ ↔ χ
pm2.61ne.2 ⊢ φ ∧ A ≠ B → ψ
pm2.61ne.3 ⊢ φ → χ
Assertion pm2.61ne ⊢ φ → ψ

Proof

Step Hyp Ref Expression
1 pm2.61ne.1 ⊢ A = B → ψ ↔ χ
2 pm2.61ne.2 ⊢ φ ∧ A ≠ B → ψ
3 pm2.61ne.3 ⊢ φ → χ
4 3 1 imbitrrid ⊢ A = B → φ → ψ
5 2 expcom ⊢ A ≠ B → φ → ψ
6 4 5 pm2.61ine ⊢ φ → ψ