Theorem neeq1 2262

Description: Equality theorem for inequality. (Contributed by NM, 19-Nov-1994.)

Assertion

Ref	Expression
neeq1	⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶))

Proof of Theorem neeq1

Step	Hyp	Ref	Expression
1		eqeq1 2089	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 = 𝐶 ↔ 𝐵 = 𝐶))
2	1	notbid 625	. 2 ⊢ (𝐴 = 𝐵 → (¬ 𝐴 = 𝐶 ↔ ¬ 𝐵 = 𝐶))
3		df-ne 2250	. 2 ⊢ (𝐴 ≠ 𝐶 ↔ ¬ 𝐴 = 𝐶)
4		df-ne 2250	. 2 ⊢ (𝐵 ≠ 𝐶 ↔ ¬ 𝐵 = 𝐶)
5	2, 3, 4	3bitr4g 221	1 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 103   = wceq 1285   ≠ wne 2249

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 577  ax-in2 578  ax-5 1377  ax-gen 1379  ax-4 1441  ax-17 1460  ax-ext 2065

This theorem depends on definitions:  df-bi 115  df-cleq 2076  df-ne 2250

This theorem is referenced by:  neeq1i  2264  neeq1d  2267  nelrdva  2806  0inp0  3960  frecabcl  6068  xnn0nemnf  8481  uzn0  8767  xrnemnf  8981  xrnepnf  8982  ngtmnft  9013  fztpval  9228