Theorem neeq12d 2532

Description: Deduction for inequality. (Contributed by NM, 24-Jul-2012.)

Hypotheses

Ref	Expression
neeq1d.1	⊢ (φ → A = B)
neeq12d.2	⊢ (φ → C = D)

Assertion

Ref	Expression
neeq12d	⊢ (φ → (A ≠ C ↔ B ≠ D))

Proof of Theorem neeq12d

Step	Hyp	Ref	Expression
1		neeq1d.1	. . 3 ⊢ (φ → A = B)
2	1	neeq1d 2530	. 2 ⊢ (φ → (A ≠ C ↔ B ≠ C))
3		neeq12d.2	. . 3 ⊢ (φ → C = D)
4	3	neeq2d 2531	. 2 ⊢ (φ → (B ≠ C ↔ B ≠ D))
5	2, 4	bitrd 244	1 ⊢ (φ → (A ≠ C ↔ B ≠ D))

This proof depends on syntax axioms:   → 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  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-11 1746  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-ex 1542  df-cleq 2346  df-ne 2519

This theorem is used by:  3netr3d  2543  3netr4d  2544