Theorem necon3abid 2550

Description: Deduction from equality to inequality. (Contributed by NM, 21-Mar-2007.)

Hypothesis

Ref	Expression
necon3abid.1	⊢ (φ → (A = B ↔ ψ))

Assertion

Ref	Expression
necon3abid	⊢ (φ → (A ≠ B ↔ ¬ ψ))

Proof of Theorem necon3abid

Step	Hyp	Ref	Expression
1		df-ne 2519	. 2 ⊢ (A ≠ B ↔ ¬ A = B)
2		necon3abid.1	. . 3 ⊢ (φ → (A = B ↔ ψ))
3	2	notbid 285	. 2 ⊢ (φ → (¬ A = B ↔ ¬ ψ))
4	1, 3	syl5bb 248	1 ⊢ (φ → (A ≠ B ↔ ¬ ψ))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   = 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-ne 2519

This theorem is used by:  necon3bbid  2551  ncpw1pwneg  6202  ltlenlec  6208