Metamath Proof Explorer


Theorem pgindlem

Description: Lemma for pgind . (Contributed by Emmett Weisz, 27-May-2024) (New usage is discouraged.)

Ref Expression
Assertion pgindlem ⊢ x ∈ 𝒫 z × 𝒫 z → 1 st ⁡ x ∪ 2 nd ⁡ x ⊆ z

Proof

Step Hyp Ref Expression
1 xp1st ⊢ x ∈ 𝒫 z × 𝒫 z → 1 st ⁡ x ∈ 𝒫 z
2 1 elpwid ⊢ x ∈ 𝒫 z × 𝒫 z → 1 st ⁡ x ⊆ z
3 xp2nd ⊢ x ∈ 𝒫 z × 𝒫 z → 2 nd ⁡ x ∈ 𝒫 z
4 3 elpwid ⊢ x ∈ 𝒫 z × 𝒫 z → 2 nd ⁡ x ⊆ z
5 2 4 unssd ⊢ x ∈ 𝒫 z × 𝒫 z → 1 st ⁡ x ∪ 2 nd ⁡ x ⊆ z