Theorem neeq12d 2440

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 2438	. 2 ⊢ (𝜑 → (𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶))
3		neeq12d.2	. . 3 ⊢ (𝜑 → 𝐶 = 𝐷)
4	3	neeq2d 2439	. 2 ⊢ (𝜑 → (𝐵 ≠ 𝐶 ↔ 𝐵 ≠ 𝐷))
5	2, 4	bitrd 188	1 ⊢ (𝜑 → (𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐷))

This proof depends on syntax axioms:   → wi 4   ↔ wb 105   = wceq 1402   ≠ wne 2420

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-5 1500  ax-gen 1502  ax-4 1563  ax-17 1579  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-cleq 2231  df-ne 2421

This theorem is used by:  3netr3d  2452  3netr4d  2453  suppval  6477  exmidapne  7626  ennnfonelemim  13315  ctinfom  13319  isnzr  14488  opprnzrbg  14492  apdiff  17097