Metamath Proof Explorer


Theorem symdifeq1

Description: Equality theorem for symmetric difference. (Contributed by Scott Fenton, 24-Apr-2012)

Ref Expression
Assertion symdifeq1 ( 𝐴 = 𝐵 → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )

Proof

Step Hyp Ref Expression
1 difeq1 ( 𝐴 = 𝐵 → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )
2 difeq2 ( 𝐴 = 𝐵 → ( 𝐶𝐴 ) = ( 𝐶𝐵 ) )
3 1 2 uneq12d ( 𝐴 = 𝐵 → ( ( 𝐴𝐶 ) ∪ ( 𝐶𝐴 ) ) = ( ( 𝐵𝐶 ) ∪ ( 𝐶𝐵 ) ) )
4 df-symdif ( 𝐴𝐶 ) = ( ( 𝐴𝐶 ) ∪ ( 𝐶𝐴 ) )
5 df-symdif ( 𝐵𝐶 ) = ( ( 𝐵𝐶 ) ∪ ( 𝐶𝐵 ) )
6 3 4 5 3eqtr4g ( 𝐴 = 𝐵 → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )