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Metamath Proof Explorer

Theorem fneq2i

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

Ref Expression
Hypothesis fneq2i.1 ⊢ A = B
Assertion fneq2i ⊢ F Fn A ↔ F Fn B

Proof

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