Theorem neeq1d 2418

Description: Deduction for inequality. (Contributed by NM, 25-Oct-1999.)

Hypothesis

Ref	Expression
neeq1d.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
neeq1d	⊢ (𝜑 → (𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶))

Proof of Theorem neeq1d

Step	Hyp	Ref	Expression
1		neeq1d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		neeq1 2413	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1395   ≠ wne 2400

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 617  ax-in2 618  ax-5 1493  ax-gen 1495  ax-4 1556  ax-17 1572  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-cleq 2222  df-ne 2401

This theorem is referenced by:  neeq12d  2420  eqnetrd  2424  prnzg  3792  pw2f1odclem  7008  hashprg  11048  algcvg  12591  algcvga  12594  eucalgcvga  12601  rpdvds  12642  phibndlem  12759  dfphi2  12763  pcaddlem  12883  ennnfoneleminc  13003  ennnfonelemex  13006  ennnfonelemhom  13007  ennnfonelemnn0  13014  ennnfonelemr  13015  ennnfonelemim  13016  ctinfomlemom  13019  setscomd  13094  lgsne0  15738  umgr2cwwkdifex  16193  dceqnconst  16542  dcapnconst  16543  nconstwlpolem  16547