Theorem fneq2i 4937

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 4931	. 2 ⊢ (A = B → (𝐹 Fn A ↔ 𝐹 Fn B))
3	1, 2	ax-mp 7	1 ⊢ (𝐹 Fn A ↔ 𝐹 Fn B)

Syntax hints:   ↔ wb 98   = wceq 1242   Fn wfn 4840

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-4 1397  ax-17 1416  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-cleq 2030  df-fn 4848

This theorem is referenced by:  fnunsn  4949  tpos0  5830