Metamath Proof Explorer


Theorem elpglem2

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

Ref Expression
Assertion elpglem2 ( ( ( 1st𝐴 ) ⊆ Pg ∧ ( 2nd𝐴 ) ⊆ Pg ) → ∃ 𝑥 ( 𝑥 ⊆ Pg ∧ ( ( 1st𝐴 ) ∈ 𝒫 𝑥 ∧ ( 2nd𝐴 ) ∈ 𝒫 𝑥 ) ) )

Proof

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