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

let A be non empty Subset of G; :: thesis: for N being Subgroup of G holds N ~ (N ~ A) = N ` (N ~ A)
let N be Subgroup of G; :: thesis: N ~ (N ~ A) = N ` (N ~ A)
thus N ~ (N ~ A) c= N ` (N ~ A) :: according to XBOOLE_0:def 10 :: thesis: N ` (N ~ A) c= N ~ (N ~ A)
proof
let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in N ~ (N ~ A) or x in N ` (N ~ A) )
assume A1: x in N ~ (N ~ A) ; :: thesis: x in N ` (N ~ A)
then reconsider x = x as Element of G ;
x * N meets N ~ A by ;
then consider z being object such that
A2: ( z in x * N & z in N ~ A ) by XBOOLE_0:3;
reconsider z = z as Element of G by A2;
z * N meets A by ;
then A3: x * N meets A by ;
x * N c= N ~ A
proof
let y be object ; :: according to TARSKI:def 3 :: thesis: ( not y in x * N or y in N ~ A )
assume A4: y in x * N ; :: thesis: y in N ~ A
then reconsider y = y as Element of G ;
x * N = y * N by ;
hence y in N ~ A by A3; :: thesis: verum
end;
hence x in N ` (N ~ A) ; :: thesis: verum
end;
let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in N ` (N ~ A) or x in N ~ (N ~ A) )
assume A5: x in N ` (N ~ A) ; :: thesis: x in N ~ (N ~ A)
then reconsider x = x as Element of G ;
A6: x * N c= N ~ A by ;
x in x * N by GROUP_2:108;
then x * N meets N ~ A by ;
hence x in N ~ (N ~ A) ; :: thesis: verum