Metamath Proof Explorer


Theorem funeq

Description: Equality theorem for function predicate. (Contributed by NM, 16-Aug-1994)

Ref Expression
Assertion funeq ( 𝐴 = 𝐵 → ( Fun 𝐴 ↔ Fun 𝐵 ) )

Proof

Step Hyp Ref Expression
1 eqimss2 ( 𝐴 = 𝐵𝐵𝐴 )
2 funss ( 𝐵𝐴 → ( Fun 𝐴 → Fun 𝐵 ) )
3 1 2 syl ( 𝐴 = 𝐵 → ( Fun 𝐴 → Fun 𝐵 ) )
4 eqimss ( 𝐴 = 𝐵𝐴𝐵 )
5 funss ( 𝐴𝐵 → ( Fun 𝐵 → Fun 𝐴 ) )
6 4 5 syl ( 𝐴 = 𝐵 → ( Fun 𝐵 → Fun 𝐴 ) )
7 3 6 impbid ( 𝐴 = 𝐵 → ( Fun 𝐴 ↔ Fun 𝐵 ) )