Metamath Proof Explorer


Theorem dffn5

Description: Representation of a function in terms of its values. (Contributed by FL, 14-Sep-2013) (Proof shortened by Mario Carneiro, 31-Aug-2015)

Ref Expression
Assertion dffn5 F Fn A F = x A F x

Proof

Step Hyp Ref Expression
1 fnrel F Fn A Rel F
2 dfrel4v Rel F F = x y | x F y
3 1 2 sylib F Fn A F = x y | x F y
4 fnbr F Fn A x F y x A
5 4 ex F Fn A x F y x A
6 5 pm4.71rd F Fn A x F y x A x F y
7 eqcom y = F x F x = y
8 fnbrfvb F Fn A x A F x = y x F y
9 7 8 syl5bb F Fn A x A y = F x x F y
10 9 pm5.32da F Fn A x A y = F x x A x F y
11 6 10 bitr4d F Fn A x F y x A y = F x
12 11 opabbidv F Fn A x y | x F y = x y | x A y = F x
13 3 12 eqtrd F Fn A F = x y | x A y = F x
14 df-mpt x A F x = x y | x A y = F x
15 13 14 eqtr4di F Fn A F = x A F x
16 fvex F x V
17 eqid x A F x = x A F x
18 16 17 fnmpti x A F x Fn A
19 fneq1 F = x A F x F Fn A x A F x Fn A
20 18 19 mpbiri F = x A F x F Fn A
21 15 20 impbii F Fn A F = x A F x