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