Metamath Proof Explorer


Theorem dfac5lem3

Description: Lemma for dfac5 . (Contributed by NM, 12-Apr-2004)

Ref Expression
Hypothesis dfac5lem.1 A = u | u t h u = t × t
Assertion dfac5lem3 w × w A w w h

Proof

Step Hyp Ref Expression
1 dfac5lem.1 A = u | u t h u = t × t
2 snex w V
3 vex w V
4 2 3 xpex w × w V
5 neeq1 u = w × w u w × w
6 eqeq1 u = w × w u = t × t w × w = t × t
7 6 rexbidv u = w × w t h u = t × t t h w × w = t × t
8 5 7 anbi12d u = w × w u t h u = t × t w × w t h w × w = t × t
9 4 8 elab w × w u | u t h u = t × t w × w t h w × w = t × t
10 1 eleq2i w × w A w × w u | u t h u = t × t
11 xpeq2 w = w × w = w ×
12 xp0 w × =
13 11 12 eqtrdi w = w × w =
14 rneq w × w = ran w × w = ran
15 3 snnz w
16 rnxp w ran w × w = w
17 15 16 ax-mp ran w × w = w
18 rn0 ran =
19 14 17 18 3eqtr3g w × w = w =
20 13 19 impbii w = w × w =
21 20 necon3bii w w × w
22 df-rex t h w × w = t × t t t h w × w = t × t
23 rneq w × w = t × t ran w × w = ran t × t
24 vex t V
25 24 snnz t
26 rnxp t ran t × t = t
27 25 26 ax-mp ran t × t = t
28 23 17 27 3eqtr3g w × w = t × t w = t
29 sneq w = t w = t
30 29 xpeq1d w = t w × w = t × w
31 xpeq2 w = t t × w = t × t
32 30 31 eqtrd w = t w × w = t × t
33 28 32 impbii w × w = t × t w = t
34 equcom w = t t = w
35 33 34 bitri w × w = t × t t = w
36 35 anbi1ci t h w × w = t × t t = w t h
37 36 exbii t t h w × w = t × t t t = w t h
38 elequ1 t = w t h w h
39 38 equsexvw t t = w t h w h
40 22 37 39 3bitrri w h t h w × w = t × t
41 21 40 anbi12i w w h w × w t h w × w = t × t
42 9 10 41 3bitr4i w × w A w w h