Metamath Proof Explorer


Theorem elpglem1

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

Ref Expression
Assertion elpglem1 ⊢ ∃ x x ⊆ Pg ∧ 1 st ⁡ A ∈ 𝒫 x ∧ 2 nd ⁡ A ∈ 𝒫 x → 1 st ⁡ A ⊆ Pg ∧ 2 nd ⁡ A ⊆ Pg

Proof

Step Hyp Ref Expression
1 elpwi ⊢ 1 st ⁡ A ∈ 𝒫 x → 1 st ⁡ A ⊆ x
2 1 adantl ⊢ x ⊆ Pg ∧ 1 st ⁡ A ∈ 𝒫 x → 1 st ⁡ A ⊆ x
3 simpl ⊢ x ⊆ Pg ∧ 1 st ⁡ A ∈ 𝒫 x → x ⊆ Pg
4 2 3 sstrd ⊢ x ⊆ Pg ∧ 1 st ⁡ A ∈ 𝒫 x → 1 st ⁡ A ⊆ Pg
5 elpwi ⊢ 2 nd ⁡ A ∈ 𝒫 x → 2 nd ⁡ A ⊆ x
6 5 adantl ⊢ x ⊆ Pg ∧ 2 nd ⁡ A ∈ 𝒫 x → 2 nd ⁡ A ⊆ x
7 simpl ⊢ x ⊆ Pg ∧ 2 nd ⁡ A ∈ 𝒫 x → x ⊆ Pg
8 6 7 sstrd ⊢ x ⊆ Pg ∧ 2 nd ⁡ A ∈ 𝒫 x → 2 nd ⁡ A ⊆ Pg
9 4 8 anim12dan ⊢ x ⊆ Pg ∧ 1 st ⁡ A ∈ 𝒫 x ∧ 2 nd ⁡ A ∈ 𝒫 x → 1 st ⁡ A ⊆ Pg ∧ 2 nd ⁡ A ⊆ Pg
10 9 exlimiv ⊢ ∃ x x ⊆ Pg ∧ 1 st ⁡ A ∈ 𝒫 x ∧ 2 nd ⁡ A ∈ 𝒫 x → 1 st ⁡ A ⊆ Pg ∧ 2 nd ⁡ A ⊆ Pg