Metamath Proof Explorer


Theorem fnbrfvb

Description: Equivalence of function value and binary relation. (Contributed by NM, 19-Apr-2004) (Revised by Mario Carneiro, 28-Apr-2015)

Ref Expression
Assertion fnbrfvb F Fn A B A F B = C B F C

Proof

Step Hyp Ref Expression
1 eqid F B = F B
2 fvex F B V
3 eqeq2 x = F B F B = x F B = F B
4 breq2 x = F B B F x B F F B
5 3 4 bibi12d x = F B F B = x B F x F B = F B B F F B
6 5 imbi2d x = F B F Fn A B A F B = x B F x F Fn A B A F B = F B B F F B
7 fneu F Fn A B A ∃! x B F x
8 tz6.12c ∃! x B F x F B = x B F x
9 7 8 syl F Fn A B A F B = x B F x
10 2 6 9 vtocl F Fn A B A F B = F B B F F B
11 1 10 mpbii F Fn A B A B F F B
12 breq2 F B = C B F F B B F C
13 11 12 syl5ibcom F Fn A B A F B = C B F C
14 fnfun F Fn A Fun F
15 funbrfv Fun F B F C F B = C
16 14 15 syl F Fn A B F C F B = C
17 16 adantr F Fn A B A B F C F B = C
18 13 17 impbid F Fn A B A F B = C B F C