Metamath Proof Explorer


Theorem fpwwe2lem1

Description: Lemma for fpwwe2 . (Contributed by Mario Carneiro, 15-May-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 fpwwe2lem1 W 𝒫 A × 𝒫 A × A

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 simpll x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y x A
3 velpw x 𝒫 A x A
4 2 3 sylibr x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y x 𝒫 A
5 simplr x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y r x × x
6 xpss12 x A x A x × x A × A
7 2 2 6 syl2anc x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y x × x A × A
8 5 7 sstrd x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y r A × A
9 velpw r 𝒫 A × A r A × A
10 8 9 sylibr x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y r 𝒫 A × A
11 4 10 jca x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y x 𝒫 A r 𝒫 A × A
12 11 ssopab2i x r | x A r x × x r We x y x [˙ r -1 y / u]˙ u F r u × u = y x r | x 𝒫 A r 𝒫 A × A
13 df-xp 𝒫 A × 𝒫 A × A = x r | x 𝒫 A r 𝒫 A × A
14 12 1 13 3sstr4i W 𝒫 A × 𝒫 A × A