Metamath Proof Explorer


Theorem fpwwe2cbv

Description: Lemma for fpwwe2 . (Contributed by Mario Carneiro, 3-Jun-2015)

Ref Expression
Hypothesis fpwwe2.1 W = x r | x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y
Assertion fpwwe2cbv W = a s | a A s a × a s We a z a [˙ s -1 z / v]˙ v F s v × v = z

Proof

Step Hyp Ref Expression
1 fpwwe2.1 W = x r | x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y
2 simpl x = a r = s x = a
3 2 sseq1d x = a r = s x A a A
4 simpr x = a r = s r = s
5 2 sqxpeqd x = a r = s x × x = a × a
6 4 5 sseq12d x = a r = s r x × x s a × a
7 3 6 anbi12d x = a r = s x A r x × x a A s a × a
8 4 2 weeq12d x = a r = s r We x s We a
9 id u = v u = v
10 9 sqxpeqd u = v u × u = v × v
11 10 ineq2d u = v r u × u = r v × v
12 9 11 oveq12d u = v u F r u × u = v F r v × v
13 12 eqeq1d u = v u F r u × u = y v F r v × v = y
14 13 cbvsbcvw [˙ r -1 y / u]˙ u F r u × u = y [˙ r -1 y / v]˙ v F r v × v = y
15 sneq y = z y = z
16 15 imaeq2d y = z r -1 y = r -1 z
17 eqeq2 y = z v F r v × v = y v F r v × v = z
18 16 17 sbceqbid y = z [˙ r -1 y / v]˙ v F r v × v = y [˙ r -1 z / v]˙ v F r v × v = z
19 14 18 bitrid y = z [˙ r -1 y / u]˙ u F r u × u = y [˙ r -1 z / v]˙ v F r v × v = z
20 19 cbvralvw y x [˙ r -1 y / u]˙ u F r u × u = y z x [˙ r -1 z / v]˙ v F r v × v = z
21 4 cnveqd x = a r = s r -1 = s -1
22 21 imaeq1d x = a r = s r -1 z = s -1 z
23 4 ineq1d x = a r = s r v × v = s v × v
24 23 oveq2d x = a r = s v F r v × v = v F s v × v
25 24 eqeq1d x = a r = s v F r v × v = z v F s v × v = z
26 22 25 sbceqbid x = a r = s [˙ r -1 z / v]˙ v F r v × v = z [˙ s -1 z / v]˙ v F s v × v = z
27 2 26 raleqbidv x = a r = s z x [˙ r -1 z / v]˙ v F r v × v = z z a [˙ s -1 z / v]˙ v F s v × v = z
28 20 27 bitrid x = a r = s y x [˙ r -1 y / u]˙ u F r u × u = y z a [˙ s -1 z / v]˙ v F s v × v = z
29 8 28 anbi12d x = a r = s r We x y x [˙ r -1 y / u]˙ u F r u × u = y s We a z a [˙ s -1 z / v]˙ v F s v × v = z
30 7 29 anbi12d x = a r = s x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y a A s a × a s We a z a [˙ s -1 z / v]˙ v F s v × v = z
31 30 cbvopabv x r | x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y = a s | a A s a × a s We a z a [˙ s -1 z / v]˙ v F s v × v = z
32 1 31 eqtri W = a s | a A s a × a s We a z a [˙ s -1 z / v]˙ v F s v × v = z