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 set ; :: according to TARSKI:def 3 :: thesis: ( not x in N ` (N ~ (N ` A)) or x in N ` A )
assume x in N ` (N ~ (N ` A)) ; :: thesis: x in N ` A
then consider x1 being Element of G such that
A1: ( x1 = x & x1 * N c= N ~ (N ` A) ) ;
x1 in x1 * N by GROUP_2:108;
then x1 * N meets N ` A by A1, Th33;
then consider y being set such that
A2: ( y in x1 * N & y in N ` A ) by XBOOLE_0:3;
reconsider y = y as Element of G by A2;
y * N c= A by A2, Th12;
then x1 * N c= A by A2, Th2;
hence x in N ` A by A1; :: thesis: verum

end;

let x be set ; :: according to TARSKI:def 3 :: thesis: ( not x in N ` A or x in N ` (N ~ (N ` A)) )
assume A3: x in N ` A ; :: thesis: x in N ` (N ~ (N ` A))
then reconsider x = x as Element of G ;
x * N c= N ~ (N ` A)

proof

let y be set ; :: according to TARSKI:def 3 :: thesis: ( not y in x * N or y in N ~ (N ` A) )
assume A4: y in x * N ; :: thesis: y in N ~ (N ` A)
then reconsider y = y as Element of G ;
y * N = x * N by A4, Th2;
then x in y * N by GROUP_2:108;
then y * N meets N ` A by A3, XBOOLE_0:3;
hence y in N ~ (N ` A) ; :: thesis: verum

end;

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