Theorem disjss1

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

	Ref	Expression
Assertion	disjss1	|- ( A C_ B -> ( Disj_ x e. B C -> Disj_ x e. A C ) )

Proof

Step	Hyp	Ref	Expression
1		ssel	|- ( A C_ B -> ( x e. A -> x e. B ) )
2	1	anim1d	|- ( A C_ B -> ( ( x e. A /\ y e. C ) -> ( x e. B /\ y e. C ) ) )
3	2	moimdv	|- ( A C_ B -> ( E* x ( x e. B /\ y e. C ) -> E* x ( x e. A /\ y e. C ) ) )
4	3	alimdv	|- ( A C_ B -> ( A. y E* x ( x e. B /\ y e. C ) -> A. y E* x ( x e. A /\ y e. C ) ) )
5		dfdisj2	|- ( Disj_ x e. B C <-> A. y E* x ( x e. B /\ y e. C ) )
6		dfdisj2	|- ( Disj_ x e. A C <-> A. y E* x ( x e. A /\ y e. C ) )
7	4 5 6	3imtr4g	|- ( A C_ B -> ( Disj_ x e. B C -> Disj_ x e. A C ) )