Metamath Proof Explorer


Theorem disjxsn

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

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

Proof

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