Theorem pm2.61ne 2592

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

Hypotheses

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

Assertion

Ref	Expression
pm2.61ne	⊢ (φ → ψ)

Proof of Theorem pm2.61ne

Step	Hyp	Ref	Expression
1		pm2.61ne.2	. . 3 ⊢ ((φ ∧ A ≠ B) → ψ)
2	1	expcom 424	. 2 ⊢ (A ≠ B → (φ → ψ))
3		nne 2521	. . 3 ⊢ (¬ A ≠ B ↔ A = B)
4		pm2.61ne.3	. . . 4 ⊢ (φ → χ)
5		pm2.61ne.1	. . . 4 ⊢ (A = B → (ψ ↔ χ))
6	4, 5	syl5ibr 212	. . 3 ⊢ (A = B → (φ → ψ))
7	3, 6	sylbi 187	. 2 ⊢ (¬ A ≠ B → (φ → ψ))
8	2, 7	pm2.61i 156	1 ⊢ (φ → ψ)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∧ wa 358   = wceq 1642   ≠ wne 2517

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-an 360  df-ne 2519

This theorem is used by: (None)