Theorem fneq2d 5418

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 5416	. 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 1395   Fn wfn 5319

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 1493  ax-gen 1495  ax-4 1556  ax-17 1572  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-cleq 2222  df-fn 5327

This theorem is referenced by:  fneq12d  5419  fncofn  5827  acfun  7415  ccfunen  7476  ccatlid  11176  ccatrid  11177  ccatass  11178  ccatswrd  11244  swrdccat2  11245  ccatpfx  11275  swrdswrd  11279  swrdccatin2  11303  pfxccatin12  11307  seq3shft  11392  ptex  13340  srg1zr  13993