Metamath Proof Explorer


Theorem elsymdifxor

Description: Membership in a symmetric difference is an exclusive-or relationship. (Contributed by David A. Wheeler, 26-Apr-2020) (Proof shortened by BJ, 13-Aug-2022)

Ref Expression
Assertion elsymdifxor
|- ( A e. ( B /_\ C ) <-> ( A e. B \/_ A e. C ) )

Proof

Step Hyp Ref Expression
1 elsymdif
 |-  ( A e. ( B /_\ C ) <-> -. ( A e. B <-> A e. C ) )
2 df-xor
 |-  ( ( A e. B \/_ A e. C ) <-> -. ( A e. B <-> A e. C ) )
3 1 2 bitr4i
 |-  ( A e. ( B /_\ C ) <-> ( A e. B \/_ A e. C ) )