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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 )