Metamath Proof Explorer


Theorem feq23

Description: Equality theorem for functions. (Contributed by FL, 14-Jul-2007) (Proof shortened by Andrew Salmon, 17-Sep-2011)

Ref Expression
Assertion feq23 A = C B = D F : A B F : C D

Proof

Step Hyp Ref Expression
1 feq2 A = C F : A B F : C B
2 feq3 B = D F : C B F : C D
3 1 2 sylan9bb A = C B = D F : A B F : C D