Metamath Proof Explorer


Theorem feq1

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

Ref Expression
Assertion feq1
|- ( F = G -> ( F : A --> B <-> G : A --> B ) )

Proof

Step Hyp Ref Expression
1 fneq1
 |-  ( F = G -> ( F Fn A <-> G Fn A ) )
2 rneq
 |-  ( F = G -> ran F = ran G )
3 2 sseq1d
 |-  ( F = G -> ( ran F C_ B <-> ran G C_ B ) )
4 1 3 anbi12d
 |-  ( F = G -> ( ( F Fn A /\ ran F C_ B ) <-> ( G Fn A /\ ran G C_ B ) ) )
5 df-f
 |-  ( F : A --> B <-> ( F Fn A /\ ran F C_ B ) )
6 df-f
 |-  ( G : A --> B <-> ( G Fn A /\ ran G C_ B ) )
7 4 5 6 3bitr4g
 |-  ( F = G -> ( F : A --> B <-> G : A --> B ) )