Metamath Proof Explorer


Theorem disjss1

Description: A subset of a disjoint collection is disjoint. (Contributed by Mario Carneiro, 14-Nov-2016)

Ref Expression
Assertion disjss1 ( 𝐴𝐵 → ( Disj 𝑥𝐵 𝐶Disj 𝑥𝐴 𝐶 ) )

Proof

Step Hyp Ref Expression
1 ssel ( 𝐴𝐵 → ( 𝑥𝐴𝑥𝐵 ) )
2 1 anim1d ( 𝐴𝐵 → ( ( 𝑥𝐴𝑦𝐶 ) → ( 𝑥𝐵𝑦𝐶 ) ) )
3 2 moimdv ( 𝐴𝐵 → ( ∃* 𝑥 ( 𝑥𝐵𝑦𝐶 ) → ∃* 𝑥 ( 𝑥𝐴𝑦𝐶 ) ) )
4 3 alimdv ( 𝐴𝐵 → ( ∀ 𝑦 ∃* 𝑥 ( 𝑥𝐵𝑦𝐶 ) → ∀ 𝑦 ∃* 𝑥 ( 𝑥𝐴𝑦𝐶 ) ) )
5 dfdisj2 ( Disj 𝑥𝐵 𝐶 ↔ ∀ 𝑦 ∃* 𝑥 ( 𝑥𝐵𝑦𝐶 ) )
6 dfdisj2 ( Disj 𝑥𝐴 𝐶 ↔ ∀ 𝑦 ∃* 𝑥 ( 𝑥𝐴𝑦𝐶 ) )
7 4 5 6 3imtr4g ( 𝐴𝐵 → ( Disj 𝑥𝐵 𝐶Disj 𝑥𝐴 𝐶 ) )