Metamath Proof Explorer


Theorem disjss3

Description: Expand a disjoint collection with any number of empty sets. (Contributed by Mario Carneiro, 15-Nov-2016)

Ref Expression
Assertion disjss3 A B x B A C = Disj x A C Disj x B C

Proof

Step Hyp Ref Expression
1 df-ral x B A C = x x B A C =
2 simprr x B A C = x B y C y C
3 n0i y C ¬ C =
4 2 3 syl x B A C = x B y C ¬ C =
5 simpl x B y C x B
6 5 adantl x B A C = x B y C x B
7 eldif x B A x B ¬ x A
8 simpl x B A C = x B y C x B A C =
9 7 8 syl5bir x B A C = x B y C x B ¬ x A C =
10 6 9 mpand x B A C = x B y C ¬ x A C =
11 4 10 mt3d x B A C = x B y C x A
12 11 2 jca x B A C = x B y C x A y C
13 12 ex x B A C = x B y C x A y C
14 13 alimi x x B A C = x x B y C x A y C
15 1 14 sylbi x B A C = x x B y C x A y C
16 moim x x B y C x A y C * x x A y C * x x B y C
17 15 16 syl x B A C = * x x A y C * x x B y C
18 17 alimdv x B A C = y * x x A y C y * x x B y C
19 dfdisj2 Disj x A C y * x x A y C
20 dfdisj2 Disj x B C y * x x B y C
21 18 19 20 3imtr4g x B A C = Disj x A C Disj x B C
22 21 adantl A B x B A C = Disj x A C Disj x B C
23 disjss1 A B Disj x B C Disj x A C
24 23 adantr A B x B A C = Disj x B C Disj x A C
25 22 24 impbid A B x B A C = Disj x A C Disj x B C