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Metamath Proof Explorer

Theorem unissd

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

		Ref	Expression
	Hypothesis	unissd.1	$⊢ φ \to A \subseteq B$
	Assertion	unissd	$⊢ φ \to ⋃ A \subseteq ⋃ B$

Step	Hyp	Ref	Expression
1		unissd.1	$⊢ φ \to A \subseteq B$
2		uniss	$⊢ A \subseteq B \to ⋃ A \subseteq ⋃ B$
3	1 2	syl	$⊢ φ \to ⋃ A \subseteq ⋃ B$