Theorem necon3bid 2297

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 2257	. 2 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵)
2		necon3bid.1	. . 3 ⊢ (𝜑 → (𝐴 = 𝐵 ↔ 𝐶 = 𝐷))
3	2	necon3bbid 2296	. 2 ⊢ (𝜑 → (¬ 𝐴 = 𝐵 ↔ 𝐶 ≠ 𝐷))
4	1, 3	syl5bb 191	1 ⊢ (𝜑 → (𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 104   = wceq 1290   ≠ wne 2256

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 580  ax-in2 581

This theorem depends on definitions:  df-bi 116  df-ne 2257

This theorem is referenced by:  nebidc  2336  addneintrd  7733  addneintr2d  7734  negne0bd  7849  negned  7853  subne0d  7865  subne0ad  7867  subneintrd  7900  subneintr2d  7902  qapne  9187  xrlttri3  9330  sqne0  10083  fihashneq0  10266  hashnncl  10267  cjne0  10405  absne0d  10683  sqrt2irraplemnn  11498