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  7003  hashprg  11038  algcvg  12578  algcvga  12581  eucalgcvga  12588  rpdvds  12629  phibndlem  12746  dfphi2  12750  pcaddlem  12870  ennnfoneleminc  12990  ennnfonelemex  12993  ennnfonelemhom  12994  ennnfonelemnn0  13001  ennnfonelemr  13002  ennnfonelemim  13003  ctinfomlemom  13006  setscomd  13081  lgsne0  15725  dceqnconst  16458  dcapnconst  16459  nconstwlpolem  16463