Theorem neeq1d 2438

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 2433	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐴 ≠ 𝐶 ↔ 𝐵 ≠ 𝐶))

This proof depends on syntax axioms:   → 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  ax-5 1500  ax-gen 1502  ax-4 1563  ax-17 1579  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-cleq 2231  df-ne 2421

This theorem is used by:  neeq12d  2440  eqnetrd  2444  prnzg  3838  suppval1  6479  elsuppfng  6482  elsuppfn  6483  suppsnopdc  6490  ressuppss  6494  pw2f1odclem  7134  hashprg  11251  algcvg  12828  algcvga  12831  eucalgcvga  12838  rpdvds  12879  phibndlem  12996  dfphi2  13000  pcaddlem  13120  ennnfoneleminc  13304  ennnfonelemex  13307  ennnfonelemhom  13308  ennnfonelemnn0  13315  ennnfonelemr  13316  ennnfonelemim  13317  ctinfomlemom  13320  setscomd  13395  rrgsupp  14576  pellexlem3  16099  lgsne0  16169  umgr2cwwkdifex  16678  dceqnconst  17122  dcapnconst  17123  nconstwlpolem  17127