let G be Group; :: thesis: for A being non empty Subset of G
for N being Subgroup of G holds N ~ (N ` (N ~ A)) = N ~ A
let A be non empty Subset of G; :: thesis: for N being Subgroup of G holds N ~ (N ` (N ~ A)) = N ~ A
let N be Subgroup of G; :: thesis: N ~ (N ` (N ~ A)) = N ~ A
thus N ~ (N ` (N ~ A)) c= N ~ A :: according to XBOOLE_0:def 10 :: thesis: N ~ A c= N ~ (N ` (N ~ A))

proof

let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in N ~ (N ` (N ~ A)) or x in N ~ A )
assume A1: x in N ~ (N ` (N ~ A)) ; :: thesis: x in N ~ A
then reconsider x = x as Element of G ;
x * N meets N ` (N ~ A) by A1, Th30;
then consider y being object such that
A2: ( y in x * N & y in N ` (N ~ A) ) by XBOOLE_0:3;
reconsider y = y as Element of G by A2;
y * N c= N ~ A by A2, Th31;
then A3: x * N c= N ~ A by A2, Th2;
x in x * N by GROUP_2:108;
hence x in N ~ A by A3; :: thesis: verum

end;

thus N ~ A c= N ~ (N ` (N ~ A)) :: thesis: verum

proof

A c= N ` (N ~ A)

proof

let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in A or x in N ` (N ~ A) )
assume A4: x in A ; :: thesis: x in N ` (N ~ A)
then reconsider x = x as Element of G ;
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 A5: y in x * N ; :: thesis: y in N ~ A
then reconsider y = y as Element of G ;
y * N = x * N by A5, Th2;
then x in y * N by GROUP_2:108;
then y * N meets A by A4, XBOOLE_0:3;
hence y in N ~ A ; :: thesis: verum

end;

hence x in N ` (N ~ A) ; :: thesis: verum

end;

hence N ~ A c= N ~ (N ` (N ~ A)) by Th39; :: thesis: verum

end;