Metamath Proof Explorer


Theorem dfrn5

Description: Definition of range in terms of 2nd and image. (Contributed by Scott Fenton, 17-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014) (Proof shortened by Peter Mazsa, 2-Oct-2022)

Ref Expression
Assertion dfrn5 ran A = 2 nd V × V A

Proof

Step Hyp Ref Expression
1 excom y p z p = y z p 2 nd x p A p y z p = y z p 2 nd x p A
2 opex y z V
3 breq1 p = y z p 2 nd x y z 2 nd x
4 eleq1 p = y z p A y z A
5 3 4 anbi12d p = y z p 2 nd x p A y z 2 nd x y z A
6 vex y V
7 vex z V
8 6 7 br2ndeq y z 2 nd x x = z
9 equcom x = z z = x
10 8 9 bitri y z 2 nd x z = x
11 10 anbi1i y z 2 nd x y z A z = x y z A
12 5 11 bitrdi p = y z p 2 nd x p A z = x y z A
13 2 12 ceqsexv p p = y z p 2 nd x p A z = x y z A
14 13 exbii z p p = y z p 2 nd x p A z z = x y z A
15 excom z p p = y z p 2 nd x p A p z p = y z p 2 nd x p A
16 vex x V
17 opeq2 z = x y z = y x
18 17 eleq1d z = x y z A y x A
19 16 18 ceqsexv z z = x y z A y x A
20 14 15 19 3bitr3ri y x A p z p = y z p 2 nd x p A
21 20 exbii y y x A y p z p = y z p 2 nd x p A
22 ancom p A p 2 nd V × V x p 2 nd V × V x p A
23 anass y z p = y z p 2 nd x p A y z p = y z p 2 nd x p A
24 16 brresi p 2 nd V × V x p V × V p 2 nd x
25 elvv p V × V y z p = y z
26 25 anbi1i p V × V p 2 nd x y z p = y z p 2 nd x
27 24 26 bitri p 2 nd V × V x y z p = y z p 2 nd x
28 27 anbi1i p 2 nd V × V x p A y z p = y z p 2 nd x p A
29 19.41vv y z p = y z p 2 nd x p A y z p = y z p 2 nd x p A
30 23 28 29 3bitr4i p 2 nd V × V x p A y z p = y z p 2 nd x p A
31 22 30 bitri p A p 2 nd V × V x y z p = y z p 2 nd x p A
32 31 exbii p p A p 2 nd V × V x p y z p = y z p 2 nd x p A
33 1 21 32 3bitr4i y y x A p p A p 2 nd V × V x
34 16 elrn2 x ran A y y x A
35 16 elima2 x 2 nd V × V A p p A p 2 nd V × V x
36 33 34 35 3bitr4i x ran A x 2 nd V × V A
37 36 eqriv ran A = 2 nd V × V A