Theorem fneq2i 5476

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 5470	. 2 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))
3	1, 2	ax-mp 5	1 ⊢ (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵)

This proof depends on syntax axioms:   ↔ wb 105   = wceq 1402   Fn wfn 5372

This proof depends on 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 proof depends on definitions:  df-bi 117  df-cleq 2231  df-fn 5380

This theorem is used by:  fnunsn  5490  tpos0  6545  dfixp  6982  xnn0nnen  10874  ser0f  10971  fnpr2o  13660  vtxedgfi  16530  vtxlpfi  16531