Theorem fneq2d 5470

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 5468	. 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 1402   Fn wfn 5370

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 1500  ax-gen 1502  ax-4 1563  ax-17 1579  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-cleq 2231  df-fn 5378

This theorem is referenced by:  fneq12d  5471  fncofn  5887  acfun  7557  ccfunen  7624  ccatlid  11357  ccatrid  11358  ccatass  11359  ccatswrd  11425  swrdccat2  11426  ccatpfx  11456  swrdswrd  11460  swrdccatin2  11484  pfxccatin12  11488  seq3shft  11586  ptex  13601  rng1zrlem  14241