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=CB=DF:ABF:CD

Proof

Step Hyp Ref Expression
1 feq2 A=CF:ABF:CB
2 feq3 B=DF:CBF:CD
3 1 2 sylan9bb A=CB=DF:ABF:CD