Theorem necon3abid 2459

Description: Deduction from equality to inequality. (Contributed by NM, 21-Mar-2007.)

Hypothesis

Ref	Expression
necon3abid.1	⊢ (𝜑 → (𝐴 = 𝐵 ↔ 𝜓))

Assertion

Ref	Expression
necon3abid	⊢ (𝜑 → (𝐴 ≠ 𝐵 ↔ ¬ 𝜓))

Proof of Theorem necon3abid

Step	Hyp	Ref	Expression
1		df-ne 2421	. 2 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵)
2		necon3abid.1	. . 3 ⊢ (𝜑 → (𝐴 = 𝐵 ↔ 𝜓))
3	2	notbid 677	. 2 ⊢ (𝜑 → (¬ 𝐴 = 𝐵 ↔ ¬ 𝜓))
4	1, 3	bitrid 192	1 ⊢ (𝜑 → (𝐴 ≠ 𝐵 ↔ ¬ 𝜓))

This proof depends on syntax axioms:  ¬ wn 3   → 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

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

This theorem is used by:  necon3bbid  2460  fndmdif  5814  suppsnopdc  6490  expnegap0  10986  gcdn0gt0  12757  cncongr2  12884  mulgnegnn  13937  domnmuln0  14584