Theorem fneq2d 5421

Description: Equality deduction for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011.)

Hypothesis

Ref	Expression
fneq2d.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
fneq2d	⊢ (𝜑 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

Proof of Theorem fneq2d

Step	Hyp	Ref	Expression
1		fneq2d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		fneq2 5419	. 2 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1397   Fn wfn 5321

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-5 1495  ax-gen 1497  ax-4 1558  ax-17 1574  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-cleq 2224  df-fn 5329

This theorem is referenced by:  fneq12d  5422  fncofn  5832  acfun  7422  ccfunen  7483  ccatlid  11187  ccatrid  11188  ccatass  11189  ccatswrd  11255  swrdccat2  11256  ccatpfx  11286  swrdswrd  11290  swrdccatin2  11314  pfxccatin12  11318  seq3shft  11403  ptex  13352  srg1zr  14006