Theorem unissd

Description: Subclass relationship for subclass union. Deduction form of uniss . (Contributed by David Moews, 1-May-2017)

	Ref	Expression
Hypothesis	unissd.1	⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
Assertion	unissd	⊢ ( 𝜑 → ∪ 𝐴 ⊆ ∪ 𝐵 )

Proof

Step	Hyp	Ref	Expression
1		unissd.1	⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
2		uniss	⊢ ( 𝐴 ⊆ 𝐵 → ∪ 𝐴 ⊆ ∪ 𝐵 )
3	1 2	syl	⊢ ( 𝜑 → ∪ 𝐴 ⊆ ∪ 𝐵 )