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Metamath Proof Explorer

Theorem pm2.61dane

Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 30-Nov-2011)

Ref Expression
Hypotheses pm2.61dane.1 ⊢ φ ∧ A = B → ψ
pm2.61dane.2 ⊢ φ ∧ A ≠ B → ψ
Assertion pm2.61dane ⊢ φ → ψ

Proof

Step Hyp Ref Expression
1 pm2.61dane.1 ⊢ φ ∧ A = B → ψ
2 pm2.61dane.2 ⊢ φ ∧ A ≠ B → ψ
3 1 ex ⊢ φ → A = B → ψ
4 2 ex ⊢ φ → A ≠ B → ψ
5 3 4 pm2.61dne ⊢ φ → ψ