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 ( ( 𝐴𝐶𝐵𝐶 ) ↔ ( 𝐴𝐵 ) ⊆ 𝐶 )

Proof

Step Hyp Ref Expression
1 dfss2 ( ( 𝐴𝐵 ) ⊆ 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ ( 𝐴𝐵 ) → 𝑥𝐶 ) )
2 19.26 ( ∀ 𝑥 ( ( 𝑥𝐴𝑥𝐶 ) ∧ ( 𝑥𝐵𝑥𝐶 ) ) ↔ ( ∀ 𝑥 ( 𝑥𝐴𝑥𝐶 ) ∧ ∀ 𝑥 ( 𝑥𝐵𝑥𝐶 ) ) )
3 elunant ( ( 𝑥 ∈ ( 𝐴𝐵 ) → 𝑥𝐶 ) ↔ ( ( 𝑥𝐴𝑥𝐶 ) ∧ ( 𝑥𝐵𝑥𝐶 ) ) )
4 3 albii ( ∀ 𝑥 ( 𝑥 ∈ ( 𝐴𝐵 ) → 𝑥𝐶 ) ↔ ∀ 𝑥 ( ( 𝑥𝐴𝑥𝐶 ) ∧ ( 𝑥𝐵𝑥𝐶 ) ) )
5 dfss2 ( 𝐴𝐶 ↔ ∀ 𝑥 ( 𝑥𝐴𝑥𝐶 ) )
6 dfss2 ( 𝐵𝐶 ↔ ∀ 𝑥 ( 𝑥𝐵𝑥𝐶 ) )
7 5 6 anbi12i ( ( 𝐴𝐶𝐵𝐶 ) ↔ ( ∀ 𝑥 ( 𝑥𝐴𝑥𝐶 ) ∧ ∀ 𝑥 ( 𝑥𝐵𝑥𝐶 ) ) )
8 2 4 7 3bitr4i ( ∀ 𝑥 ( 𝑥 ∈ ( 𝐴𝐵 ) → 𝑥𝐶 ) ↔ ( 𝐴𝐶𝐵𝐶 ) )
9 1 8 bitr2i ( ( 𝐴𝐶𝐵𝐶 ) ↔ ( 𝐴𝐵 ) ⊆ 𝐶 )