Metamath Proof Explorer


Theorem feq12i

Description: Equality inference for functions. (Contributed by AV, 7-Feb-2021)

Ref Expression
Hypotheses feq12i.1
|- F = G
feq12i.2
|- A = B
Assertion feq12i
|- ( F : A --> C <-> G : B --> C )

Proof

Step Hyp Ref Expression
1 feq12i.1
 |-  F = G
2 feq12i.2
 |-  A = B
3 eqid
 |-  C = C
4 feq123
 |-  ( ( F = G /\ A = B /\ C = C ) -> ( F : A --> C <-> G : B --> C ) )
5 1 2 3 4 mp3an
 |-  ( F : A --> C <-> G : B --> C )