Theorem sndisj

Description: Any collection of singletons is disjoint. (Contributed by Mario Carneiro, 14-Nov-2016)

	Ref	Expression
Assertion	sndisj	|- Disj_ x e. A { x }

Proof

Step	Hyp	Ref	Expression
1		dfdisj2	|- ( Disj_ x e. A { x } <-> A. y E* x ( x e. A /\ y e. { x } ) )
2		moeq	|- E* x x = y
3		simpr	|- ( ( x e. A /\ y e. { x } ) -> y e. { x } )
4	3	elsnd	|- ( ( x e. A /\ y e. { x } ) -> y = x )
5	4	equcomd	|- ( ( x e. A /\ y e. { x } ) -> x = y )
6	5	moimi	|- ( E* x x = y -> E* x ( x e. A /\ y e. { x } ) )
7	2 6	ax-mp	|- E* x ( x e. A /\ y e. { x } )
8	1 7	mpgbir	|- Disj_ x e. A { x }