Metamath Proof Explorer


Theorem difsn

Description: An element not in a set can be removed without affecting the set. (Contributed by NM, 16-Mar-2006) (Proof shortened by Andrew Salmon, 29-Jun-2011)

Ref Expression
Assertion difsn
|- ( -. A e. B -> ( B \ { A } ) = B )

Proof

Step Hyp Ref Expression
1 eldifsn
 |-  ( x e. ( B \ { A } ) <-> ( x e. B /\ x =/= A ) )
2 simpl
 |-  ( ( x e. B /\ x =/= A ) -> x e. B )
3 nelelne
 |-  ( -. A e. B -> ( x e. B -> x =/= A ) )
4 3 ancld
 |-  ( -. A e. B -> ( x e. B -> ( x e. B /\ x =/= A ) ) )
5 2 4 impbid2
 |-  ( -. A e. B -> ( ( x e. B /\ x =/= A ) <-> x e. B ) )
6 1 5 bitrid
 |-  ( -. A e. B -> ( x e. ( B \ { A } ) <-> x e. B ) )
7 6 eqrdv
 |-  ( -. A e. B -> ( B \ { A } ) = B )