Theorem fneq2i 5293

Description: Equality inference for function predicate with domain. (Contributed by NM, 4-Sep-2011.)

Hypothesis

Ref	Expression
fneq2i.1	⊢ 𝐴 = 𝐵

Assertion

Ref	Expression
fneq2i	⊢ (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵)

Proof of Theorem fneq2i

Step	Hyp	Ref	Expression
1		fneq2i.1	. 2 ⊢ 𝐴 = 𝐵
2		fneq2 5287	. 2 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))
3	1, 2	ax-mp 5	1 ⊢ (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵)

Syntax hints:   ↔ wb 104   = wceq 1348   Fn wfn 5193

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1440  ax-gen 1442  ax-4 1503  ax-17 1519  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-cleq 2163  df-fn 5201

This theorem is referenced by:  fnunsn  5305  tpos0  6253  dfixp  6678  ser0f  10471