Metamath Proof Explorer


Theorem elpglem1

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

Ref Expression
Assertion elpglem1 ( ∃ 𝑥 ( 𝑥 ⊆ Pg ∧ ( ( 1st𝐴 ) ∈ 𝒫 𝑥 ∧ ( 2nd𝐴 ) ∈ 𝒫 𝑥 ) ) → ( ( 1st𝐴 ) ⊆ Pg ∧ ( 2nd𝐴 ) ⊆ Pg ) )

Proof

Step Hyp Ref Expression
1 elpwi ( ( 1st𝐴 ) ∈ 𝒫 𝑥 → ( 1st𝐴 ) ⊆ 𝑥 )
2 1 adantl ( ( 𝑥 ⊆ Pg ∧ ( 1st𝐴 ) ∈ 𝒫 𝑥 ) → ( 1st𝐴 ) ⊆ 𝑥 )
3 simpl ( ( 𝑥 ⊆ Pg ∧ ( 1st𝐴 ) ∈ 𝒫 𝑥 ) → 𝑥 ⊆ Pg )
4 2 3 sstrd ( ( 𝑥 ⊆ Pg ∧ ( 1st𝐴 ) ∈ 𝒫 𝑥 ) → ( 1st𝐴 ) ⊆ Pg )
5 elpwi ( ( 2nd𝐴 ) ∈ 𝒫 𝑥 → ( 2nd𝐴 ) ⊆ 𝑥 )
6 5 adantl ( ( 𝑥 ⊆ Pg ∧ ( 2nd𝐴 ) ∈ 𝒫 𝑥 ) → ( 2nd𝐴 ) ⊆ 𝑥 )
7 simpl ( ( 𝑥 ⊆ Pg ∧ ( 2nd𝐴 ) ∈ 𝒫 𝑥 ) → 𝑥 ⊆ Pg )
8 6 7 sstrd ( ( 𝑥 ⊆ Pg ∧ ( 2nd𝐴 ) ∈ 𝒫 𝑥 ) → ( 2nd𝐴 ) ⊆ Pg )
9 4 8 anim12dan ( ( 𝑥 ⊆ Pg ∧ ( ( 1st𝐴 ) ∈ 𝒫 𝑥 ∧ ( 2nd𝐴 ) ∈ 𝒫 𝑥 ) ) → ( ( 1st𝐴 ) ⊆ Pg ∧ ( 2nd𝐴 ) ⊆ Pg ) )
10 9 exlimiv ( ∃ 𝑥 ( 𝑥 ⊆ Pg ∧ ( ( 1st𝐴 ) ∈ 𝒫 𝑥 ∧ ( 2nd𝐴 ) ∈ 𝒫 𝑥 ) ) → ( ( 1st𝐴 ) ⊆ Pg ∧ ( 2nd𝐴 ) ⊆ Pg ) )