let G be Group; :: thesis: for A, B being non empty Subset of G
for N being Subgroup of G holds N ~ (A \/ B) = (N ~ A) \/ (N ~ B)

let A, B be non empty Subset of G; :: thesis: for N being Subgroup of G holds N ~ (A \/ B) = (N ~ A) \/ (N ~ B)
let N be Subgroup of G; :: thesis: N ~ (A \/ B) = (N ~ A) \/ (N ~ B)
thus N ~ (A \/ B) c= (N ~ A) \/ (N ~ B) :: according to XBOOLE_0:def 10 :: thesis: (N ~ A) \/ (N ~ B) c= N ~ (A \/ B)
proof
let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in N ~ (A \/ B) or x in (N ~ A) \/ (N ~ B) )
assume A1: x in N ~ (A \/ B) ; :: thesis: x in (N ~ A) \/ (N ~ B)
then reconsider x = x as Element of G ;
x * N meets A \/ B by ;
then ( x * N meets A or x * N meets B ) by XBOOLE_1:70;
then ( x in N ~ A or x in N ~ B ) ;
hence x in (N ~ A) \/ (N ~ B) by XBOOLE_0:def 3; :: thesis: verum
end;
let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in (N ~ A) \/ (N ~ B) or x in N ~ (A \/ B) )
assume A2: x in (N ~ A) \/ (N ~ B) ; :: thesis: x in N ~ (A \/ B)
then reconsider x = x as Element of G ;
( x in N ~ A or x in N ~ B ) by ;
then ( x * N meets A or x * N meets B ) by Th14;
then x * N meets A \/ B by XBOOLE_1:70;
hence x in N ~ (A \/ B) ; :: thesis: verum