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 ⊢ A ⊆ B → Disj x ∈ B C → Disj x ∈ A C

Proof

Step Hyp Ref Expression
1 ssel ⊢ A ⊆ B → x ∈ A → x ∈ B
2 1 anim1d ⊢ A ⊆ B → x ∈ A ∧ y ∈ C → x ∈ B ∧ y ∈ C
3 2 moimdv ⊢ A ⊆ B → ∃* x x ∈ B ∧ y ∈ C → ∃* x x ∈ A ∧ y ∈ C
4 3 alimdv ⊢ A ⊆ B → ∀ y ∃* x x ∈ B ∧ y ∈ C → ∀ y ∃* x x ∈ A ∧ y ∈ C
5 dfdisj2 ⊢ Disj x ∈ B C ↔ ∀ y ∃* x x ∈ B ∧ y ∈ C
6 dfdisj2 ⊢ Disj x ∈ A C ↔ ∀ y ∃* x x ∈ A ∧ y ∈ C
7 4 5 6 3imtr4g ⊢ A ⊆ B → Disj x ∈ B C → Disj x ∈ A C