Theorem necon3bid 2552

Description: Deduction from equality to inequality. (Contributed by NM, 23-Feb-2005.) (Proof shortened by Andrew Salmon, 25-May-2011.)

Hypothesis

Ref	Expression
necon3bid.1	⊢ (φ → (A = B ↔ C = D))

Assertion

Ref	Expression
necon3bid	⊢ (φ → (A ≠ B ↔ C ≠ D))

Proof of Theorem necon3bid

Step	Hyp	Ref	Expression
1		df-ne 2519	. 2 ⊢ (A ≠ B ↔ ¬ A = B)
2		necon3bid.1	. . 3 ⊢ (φ → (A = B ↔ C = D))
3	2	necon3bbid 2551	. 2 ⊢ (φ → (¬ A = B ↔ C ≠ D))
4	1, 3	syl5bb 248	1 ⊢ (φ → (A ≠ B ↔ C ≠ D))

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:  nebi  2588  nchoicelem17  6306