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Metamath Proof Explorer

Theorem disjssd

Description: Subclass theorem for disjoints, deduction version. (Contributed by Peter Mazsa, 28-Sep-2021)

		Ref	Expression
	Hypothesis	disjssd.1	$⊢ φ \to A \subseteq B$
	Assertion	disjssd	$⊢ φ \to (Disj B \to Disj A)$

Step	Hyp	Ref	Expression
1		disjssd.1	$⊢ φ \to A \subseteq B$
2		disjss	$⊢ A \subseteq B \to (Disj B \to Disj A)$
3	1 2	syl	$⊢ φ \to (Disj B \to Disj A)$