Theorem disjssd

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

	Ref	Expression
Hypothesis	disjssd.1	⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
Assertion	disjssd	⊢ ( 𝜑 → ( Disj 𝐵 → Disj 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		disjssd.1	⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 )
2		disjss	⊢ ( 𝐴 ⊆ 𝐵 → ( Disj 𝐵 → Disj 𝐴 ) )
3	1 2	syl	⊢ ( 𝜑 → ( Disj 𝐵 → Disj 𝐴 ) )