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  8139  addneintr2d  8140  negne0bd  8255  negned  8259  subne0d  8271  subne0ad  8273  subneintrd  8306  subneintr2d  8308  qapne  9633  xrlttri3  9791  xaddass2  9864  sqne0  10578  fihashneq0  10765  hashnncl  10766  cjne0  10908  absne0d  11187  sqrt2irraplemnn  12169  ringinvnz1ne0  13126  metn0  13660  lgsabs1  14222  neap0mkv  14587