This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

Theorem unss

Description: The union of two subclasses is a subclass. Theorem 27 of Suppes p. 27 and its converse. (Contributed by NM, 11-Jun-2004)

Ref Expression
Assertion unss ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) ↔ ( 𝐴 ∪ 𝐵 ) ⊆ 𝐶 )

Proof

Step Hyp Ref Expression
1 df-ss ⊢ ( ( 𝐴 ∪ 𝐵 ) ⊆ 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) → 𝑥 ∈ 𝐶 ) )
2 19.26 ⊢ ( ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) ∧ ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐶 ) ) ↔ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) ∧ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐶 ) ) )
3 elunant ⊢ ( ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) → 𝑥 ∈ 𝐶 ) ↔ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) ∧ ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐶 ) ) )
4 3 albii ⊢ ( ∀ 𝑥 ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) → 𝑥 ∈ 𝐶 ) ↔ ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) ∧ ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐶 ) ) )
5 df-ss ⊢ ( 𝐴 ⊆ 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) )
6 df-ss ⊢ ( 𝐵 ⊆ 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐶 ) )
7 5 6 anbi12i ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) ↔ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) ∧ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐶 ) ) )
8 2 4 7 3bitr4i ⊢ ( ∀ 𝑥 ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) → 𝑥 ∈ 𝐶 ) ↔ ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) )
9 1 8 bitr2i ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) ↔ ( 𝐴 ∪ 𝐵 ) ⊆ 𝐶 )