Theorem fneq2i 5349

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 5343	. 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 105   = wceq 1364   Fn wfn 5249

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 1458  ax-gen 1460  ax-4 1521  ax-17 1537  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-cleq 2186  df-fn 5257

This theorem is referenced by:  fnunsn  5361  tpos0  6327  dfixp  6754  xnn0nnen  10508  ser0f  10605  fnpr2o  12922