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		velsn	|- ( y e. { x } <-> y = x )
5	3 4	sylib	|- ( ( x e. A /\ y e. { x } ) -> y = x )
6	5	equcomd	|- ( ( x e. A /\ y e. { x } ) -> x = y )
7	6	moimi	|- ( E* x x = y -> E* x ( x e. A /\ y e. { x } ) )
8	2 7	ax-mp	|- E* x ( x e. A /\ y e. { x } )
9	1 8	mpgbir	|- Disj_ x e. A { x }