Theorem necon3bid 2388

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 2348	. 2 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵)
2		necon3bid.1	. . 3 ⊢ (𝜑 → (𝐴 = 𝐵 ↔ 𝐶 = 𝐷))
3	2	necon3bbid 2387	. 2 ⊢ (𝜑 → (¬ 𝐴 = 𝐵 ↔ 𝐶 ≠ 𝐷))
4	1, 3	bitrid 192	1 ⊢ (𝜑 → (𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 105   = wceq 1353   ≠ wne 2347

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 614  ax-in2 615

This theorem depends on definitions:  df-bi 117  df-ne 2348

This theorem is referenced by:  nebidc  2427  addneintrd  8143  addneintr2d  8144  negne0bd  8259  negned  8263  subne0d  8275  subne0ad  8277  subneintrd  8310  subneintr2d  8312  qapne  9637  xrlttri3  9795  xaddass2  9868  sqne0  10582  fihashneq0  10769  hashnncl  10770  cjne0  10912  absne0d  11191  sqrt2irraplemnn  12173  ringinvnz1ne0  13179  metn0  13771  lgsabs1  14333  neap0mkv  14698