Metamath Proof Explorer


Theorem fpwwe2lem3

Description: Lemma for fpwwe2 . (Contributed by Mario Carneiro, 19-May-2015) (Revised by AV, 20-Jul-2024)

Ref Expression
Hypotheses 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
fpwwe2.2 ⊢ φ → A ∈ V
fpwwe2lem3.4 ⊢ φ → X W R
Assertion fpwwe2lem3 ⊢ φ ∧ B ∈ X → R -1 B F R ∩ R -1 B × R -1 B = B

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 fpwwe2.2 ⊢ φ → A ∈ V
3 fpwwe2lem3.4 ⊢ φ → X W R
4 1 2 fpwwe2lem2 ⊢ φ → X W R ↔ X ⊆ A ∧ R ⊆ X × X ∧ R We X ∧ ∀ y ∈ X [˙ R -1 y / u]˙ u F R ∩ u × u = y
5 3 4 mpbid ⊢ φ → X ⊆ A ∧ R ⊆ X × X ∧ R We X ∧ ∀ y ∈ X [˙ R -1 y / u]˙ u F R ∩ u × u = y
6 5 simprrd ⊢ φ → ∀ y ∈ X [˙ R -1 y / u]˙ u F R ∩ u × u = y
7 sneq ⊢ y = B → y = B
8 7 imaeq2d ⊢ y = B → R -1 y = R -1 B
9 eqeq2 ⊢ y = B → u F R ∩ u × u = y ↔ u F R ∩ u × u = B
10 8 9 sbceqbid ⊢ y = B → [˙ R -1 y / u]˙ u F R ∩ u × u = y ↔ [˙ R -1 B / u]˙ u F R ∩ u × u = B
11 10 rspccva ⊢ ∀ y ∈ X [˙ R -1 y / u]˙ u F R ∩ u × u = y ∧ B ∈ X → [˙ R -1 B / u]˙ u F R ∩ u × u = B
12 6 11 sylan ⊢ φ ∧ B ∈ X → [˙ R -1 B / u]˙ u F R ∩ u × u = B
13 cnvimass ⊢ R -1 B ⊆ dom ⁡ R
14 1 relopabiv ⊢ Rel ⁡ W
15 14 brrelex2i ⊢ X W R → R ∈ V
16 dmexg ⊢ R ∈ V → dom ⁡ R ∈ V
17 3 15 16 3syl ⊢ φ → dom ⁡ R ∈ V
18 ssexg ⊢ R -1 B ⊆ dom ⁡ R ∧ dom ⁡ R ∈ V → R -1 B ∈ V
19 13 17 18 sylancr ⊢ φ → R -1 B ∈ V
20 id ⊢ u = R -1 B → u = R -1 B
21 20 sqxpeqd ⊢ u = R -1 B → u × u = R -1 B × R -1 B
22 21 ineq2d ⊢ u = R -1 B → R ∩ u × u = R ∩ R -1 B × R -1 B
23 20 22 oveq12d ⊢ u = R -1 B → u F R ∩ u × u = R -1 B F R ∩ R -1 B × R -1 B
24 23 eqeq1d ⊢ u = R -1 B → u F R ∩ u × u = B ↔ R -1 B F R ∩ R -1 B × R -1 B = B
25 24 sbcieg ⊢ R -1 B ∈ V → [˙ R -1 B / u]˙ u F R ∩ u × u = B ↔ R -1 B F R ∩ R -1 B × R -1 B = B
26 19 25 syl ⊢ φ → [˙ R -1 B / u]˙ u F R ∩ u × u = B ↔ R -1 B F R ∩ R -1 B × R -1 B = B
27 26 adantr ⊢ φ ∧ B ∈ X → [˙ R -1 B / u]˙ u F R ∩ u × u = B ↔ R -1 B F R ∩ R -1 B × R -1 B = B
28 12 27 mpbid ⊢ φ ∧ B ∈ X → R -1 B F R ∩ R -1 B × R -1 B = B