Theorem fneq2d 5452

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 5450	. 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 1398   Fn wfn 5352

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 1496  ax-gen 1498  ax-4 1559  ax-17 1575  ax-ext 2216

This theorem depends on definitions:  df-bi 117  df-cleq 2227  df-fn 5360

This theorem is referenced by:  fneq12d  5453  fncofn  5867  acfun  7527  ccfunen  7594  ccatlid  11319  ccatrid  11320  ccatass  11321  ccatswrd  11387  swrdccat2  11388  ccatpfx  11418  swrdswrd  11422  swrdccatin2  11446  pfxccatin12  11450  seq3shft  11548  ptex  13561  srg1zr  14215