Theorem neeq12d 2400

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 2398	. 2 ⊢ (𝜑 → (𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶))
3		neeq12d.2	. . 3 ⊢ (𝜑 → 𝐶 = 𝐷)
4	3	neeq2d 2399	. 2 ⊢ (𝜑 → (𝐵 ≠ 𝐶 ↔ 𝐵 ≠ 𝐷))
5	2, 4	bitrd 188	1 ⊢ (𝜑 → (𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐷))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1375   ≠ wne 2380

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 617  ax-in2 618  ax-5 1473  ax-gen 1475  ax-4 1536  ax-17 1552  ax-ext 2191

This theorem depends on definitions:  df-bi 117  df-cleq 2202  df-ne 2381

This theorem is referenced by:  3netr3d  2412  3netr4d  2413  exmidapne  7414  ennnfonelemim  12961  ctinfom  12965  isnzr  14110  opprnzrbg  14114  apdiff  16327