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