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Metamath Proof Explorer

Theorem unssd

Description: A deduction showing the union of two subclasses is a subclass. (Contributed by Jonathan Ben-Naim, 3-Jun-2011)

Ref Expression
Hypotheses unssd.1 ⊢ ( 𝜑 → 𝐴 ⊆ 𝐶 )
unssd.2 ⊢ ( 𝜑 → 𝐵 ⊆ 𝐶 )
Assertion unssd ( 𝜑 → ( 𝐴 ∪ 𝐵 ) ⊆ 𝐶 )

Proof

Step Hyp Ref Expression
1 unssd.1 ⊢ ( 𝜑 → 𝐴 ⊆ 𝐶 )
2 unssd.2 ⊢ ( 𝜑 → 𝐵 ⊆ 𝐶 )
3 unss ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) ↔ ( 𝐴 ∪ 𝐵 ) ⊆ 𝐶 )
4 3 biimpi ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( 𝐴 ∪ 𝐵 ) ⊆ 𝐶 )
5 1 2 4 syl2anc ⊢ ( 𝜑 → ( 𝐴 ∪ 𝐵 ) ⊆ 𝐶 )