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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 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 ) )