Theorem nebi 2588

Description: Contraposition law for inequality. (Contributed by NM, 28-Dec-2008.)

Assertion

Ref	Expression
nebi	⊢ ((A = B ↔ C = D) ↔ (A ≠ B ↔ C ≠ D))

Proof of Theorem nebi

Step	Hyp	Ref	Expression
1		id 19	. . 3 ⊢ ((A = B ↔ C = D) → (A = B ↔ C = D))
2	1	necon3bid 2552	. 2 ⊢ ((A = B ↔ C = D) → (A ≠ B ↔ C ≠ D))
3		id 19	. . 3 ⊢ ((A ≠ B ↔ C ≠ D) → (A ≠ B ↔ C ≠ D))
4	3	necon4bid 2583	. 2 ⊢ ((A ≠ B ↔ C ≠ D) → (A = B ↔ C = D))
5	2, 4	impbii 180	1 ⊢ ((A = B ↔ C = D) ↔ (A ≠ B ↔ C ≠ D))

This proof depends on syntax axioms:   ↔ 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: (None)