Theorem fneq2d 5428

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 5426	. 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 5328

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 2213

This theorem depends on definitions:  df-bi 117  df-cleq 2224  df-fn 5336

This theorem is referenced by:  fneq12d  5429  fncofn  5840  acfun  7465  ccfunen  7526  ccatlid  11230  ccatrid  11231  ccatass  11232  ccatswrd  11298  swrdccat2  11299  ccatpfx  11329  swrdswrd  11333  swrdccatin2  11357  pfxccatin12  11361  seq3shft  11459  ptex  13408  srg1zr  14062