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 )