Metamath Proof Explorer


Theorem fnbr

Description: The first argument of binary relation on a function belongs to the function's domain. (Contributed by NM, 7-May-2004)

Ref Expression
Assertion fnbr
|- ( ( F Fn A /\ B F C ) -> B e. A )

Proof

Step Hyp Ref Expression
1 fnrel
 |-  ( F Fn A -> Rel F )
2 releldm
 |-  ( ( Rel F /\ B F C ) -> B e. dom F )
3 1 2 sylan
 |-  ( ( F Fn A /\ B F C ) -> B e. dom F )
4 fndm
 |-  ( F Fn A -> dom F = A )
5 4 eleq2d
 |-  ( F Fn A -> ( B e. dom F <-> B e. A ) )
6 5 biimpa
 |-  ( ( F Fn A /\ B e. dom F ) -> B e. A )
7 3 6 syldan
 |-  ( ( F Fn A /\ B F C ) -> B e. A )