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 𝐴𝐹 Fn 𝐵 )

Proof

Step Hyp Ref Expression
1 fneq2i.1 𝐴 = 𝐵
2 fneq2 ( 𝐴 = 𝐵 → ( 𝐹 Fn 𝐴𝐹 Fn 𝐵 ) )
3 1 2 ax-mp ( 𝐹 Fn 𝐴𝐹 Fn 𝐵 )