Metamath Proof Explorer


Theorem elpglem2

Description: Lemma for elpg . (Contributed by Emmett Weisz, 28-Aug-2021)

Ref Expression
Assertion elpglem2 ⊢ 1 st ⁡ A ⊆ Pg ∧ 2 nd ⁡ A ⊆ Pg → ∃ x x ⊆ Pg ∧ 1 st ⁡ A ∈ 𝒫 x ∧ 2 nd ⁡ A ∈ 𝒫 x

Proof

Step Hyp Ref Expression
1 fvex ⊢ 1 st ⁡ A ∈ V
2 fvex ⊢ 2 nd ⁡ A ∈ V
3 1 2 unex ⊢ 1 st ⁡ A ∪ 2 nd ⁡ A ∈ V
4 3 isseti ⊢ ∃ x x = 1 st ⁡ A ∪ 2 nd ⁡ A
5 sseq1 ⊢ x = 1 st ⁡ A ∪ 2 nd ⁡ A → x ⊆ Pg ↔ 1 st ⁡ A ∪ 2 nd ⁡ A ⊆ Pg
6 unss ⊢ 1 st ⁡ A ⊆ Pg ∧ 2 nd ⁡ A ⊆ Pg ↔ 1 st ⁡ A ∪ 2 nd ⁡ A ⊆ Pg
7 5 6 bitr4di ⊢ x = 1 st ⁡ A ∪ 2 nd ⁡ A → x ⊆ Pg ↔ 1 st ⁡ A ⊆ Pg ∧ 2 nd ⁡ A ⊆ Pg
8 7 biimprd ⊢ x = 1 st ⁡ A ∪ 2 nd ⁡ A → 1 st ⁡ A ⊆ Pg ∧ 2 nd ⁡ A ⊆ Pg → x ⊆ Pg
9 ssun1 ⊢ 1 st ⁡ A ⊆ 1 st ⁡ A ∪ 2 nd ⁡ A
10 id ⊢ x = 1 st ⁡ A ∪ 2 nd ⁡ A → x = 1 st ⁡ A ∪ 2 nd ⁡ A
11 9 10 sseqtrrid ⊢ x = 1 st ⁡ A ∪ 2 nd ⁡ A → 1 st ⁡ A ⊆ x
12 vex ⊢ x ∈ V
13 12 elpw2 ⊢ 1 st ⁡ A ∈ 𝒫 x ↔ 1 st ⁡ A ⊆ x
14 11 13 sylibr ⊢ x = 1 st ⁡ A ∪ 2 nd ⁡ A → 1 st ⁡ A ∈ 𝒫 x
15 ssun2 ⊢ 2 nd ⁡ A ⊆ 1 st ⁡ A ∪ 2 nd ⁡ A
16 15 10 sseqtrrid ⊢ x = 1 st ⁡ A ∪ 2 nd ⁡ A → 2 nd ⁡ A ⊆ x
17 12 elpw2 ⊢ 2 nd ⁡ A ∈ 𝒫 x ↔ 2 nd ⁡ A ⊆ x
18 16 17 sylibr ⊢ x = 1 st ⁡ A ∪ 2 nd ⁡ A → 2 nd ⁡ A ∈ 𝒫 x
19 14 18 jca ⊢ x = 1 st ⁡ A ∪ 2 nd ⁡ A → 1 st ⁡ A ∈ 𝒫 x ∧ 2 nd ⁡ A ∈ 𝒫 x
20 8 19 jctird ⊢ x = 1 st ⁡ A ∪ 2 nd ⁡ A → 1 st ⁡ A ⊆ Pg ∧ 2 nd ⁡ A ⊆ Pg → x ⊆ Pg ∧ 1 st ⁡ A ∈ 𝒫 x ∧ 2 nd ⁡ A ∈ 𝒫 x
21 4 20 eximii ⊢ ∃ x 1 st ⁡ A ⊆ Pg ∧ 2 nd ⁡ A ⊆ Pg → x ⊆ Pg ∧ 1 st ⁡ A ∈ 𝒫 x ∧ 2 nd ⁡ A ∈ 𝒫 x
22 21 19.37iv ⊢ 1 st ⁡ A ⊆ Pg ∧ 2 nd ⁡ A ⊆ Pg → ∃ x x ⊆ Pg ∧ 1 st ⁡ A ∈ 𝒫 x ∧ 2 nd ⁡ A ∈ 𝒫 x