Theorem fneq2i 5368

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 5362	. 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 1372   Fn wfn 5265

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 1469  ax-gen 1471  ax-4 1532  ax-17 1548  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-cleq 2197  df-fn 5273

This theorem is referenced by:  fnunsn  5382  tpos0  6359  dfixp  6786  xnn0nnen  10580  ser0f  10677  fnpr2o  13113