Theorem disjss1f

Description: A subset of a disjoint collection is disjoint. (Contributed by Thierry Arnoux, 6-Apr-2017)

	Ref	Expression
Hypotheses	disjss1f.1	⊢ Ⅎ 𝑥 𝐴
	disjss1f.2	⊢ Ⅎ 𝑥 𝐵
Assertion	disjss1f	⊢ ( 𝐴 ⊆ 𝐵 → ( Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		disjss1f.1	⊢ Ⅎ 𝑥 𝐴
2		disjss1f.2	⊢ Ⅎ 𝑥 𝐵
3	1 2	ssrmof	⊢ ( 𝐴 ⊆ 𝐵 → ( ∃* 𝑥 ∈ 𝐵 𝑦 ∈ 𝐶 → ∃* 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ) )
4	3	alimdv	⊢ ( 𝐴 ⊆ 𝐵 → ( ∀ 𝑦 ∃* 𝑥 ∈ 𝐵 𝑦 ∈ 𝐶 → ∀ 𝑦 ∃* 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ) )
5		df-disj	⊢ ( Disj 𝑥 ∈ 𝐵 𝐶 ↔ ∀ 𝑦 ∃* 𝑥 ∈ 𝐵 𝑦 ∈ 𝐶 )
6		df-disj	⊢ ( Disj 𝑥 ∈ 𝐴 𝐶 ↔ ∀ 𝑦 ∃* 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 )
7	4 5 6	3imtr4g	⊢ ( 𝐴 ⊆ 𝐵 → ( Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶 ) )