Metamath Proof Explorer


Theorem feq2

Description: Equality theorem for functions. (Contributed by NM, 1-Aug-1994)

Ref Expression
Assertion feq2
|- ( A = B -> ( F : A --> C <-> F : B --> C ) )

Proof

Step Hyp Ref Expression
1 fneq2
 |-  ( A = B -> ( F Fn A <-> F Fn B ) )
2 1 anbi1d
 |-  ( A = B -> ( ( F Fn A /\ ran F C_ C ) <-> ( F Fn B /\ ran F C_ C ) ) )
3 df-f
 |-  ( F : A --> C <-> ( F Fn A /\ ran F C_ C ) )
4 df-f
 |-  ( F : B --> C <-> ( F Fn B /\ ran F C_ C ) )
5 2 3 4 3bitr4g
 |-  ( A = B -> ( F : A --> C <-> F : B --> C ) )