Metamath Proof Explorer


Theorem uniss

Description: Subclass relationship for class union. Theorem 61 of Suppes p. 39. (Contributed by NM, 22-Mar-1998) (Proof shortened by Andrew Salmon, 29-Jun-2011)

Ref Expression
Assertion uniss ( 𝐴 ⊆ 𝐵 → ∪ 𝐴 ⊆ ∪ 𝐵 )

Proof

Step Hyp Ref Expression
1 ssel ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑦 ∈ 𝐴 → 𝑦 ∈ 𝐵 ) )
2 1 anim2d ⊢ ( 𝐴 ⊆ 𝐵 → ( ( 𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐴 ) → ( 𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐵 ) ) )
3 2 eximdv ⊢ ( 𝐴 ⊆ 𝐵 → ( ∃ 𝑦 ( 𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐴 ) → ∃ 𝑦 ( 𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐵 ) ) )
4 eluni ⊢ ( 𝑥 ∈ ∪ 𝐴 ↔ ∃ 𝑦 ( 𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐴 ) )
5 eluni ⊢ ( 𝑥 ∈ ∪ 𝐵 ↔ ∃ 𝑦 ( 𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐵 ) )
6 3 4 5 3imtr4g ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑥 ∈ ∪ 𝐴 → 𝑥 ∈ ∪ 𝐵 ) )
7 6 ssrdv ⊢ ( 𝐴 ⊆ 𝐵 → ∪ 𝐴 ⊆ ∪ 𝐵 )