let G be Group; :: thesis: for A being non empty Subset of G

for N being Subgroup of G holds N ` (N ` A) = N ` A

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

let N be Subgroup of G; :: thesis: N ` (N ` A) = N ` A

thus N ` (N ` A) c= N ` A :: according to XBOOLE_0:def 10 :: thesis: N ` A c= N ` (N ` A)

assume A2: x in N ` A ; :: thesis: x in N ` (N ` A)

then reconsider x9 = x as Element of G ;

A3: x9 * N c= A by A2, Th12;

x9 * N c= N ` A

for N being Subgroup of G holds N ` (N ` A) = N ` A

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

let N be Subgroup of G; :: thesis: N ` (N ` A) = N ` A

thus N ` (N ` A) c= N ` A :: according to XBOOLE_0:def 10 :: thesis: N ` A c= N ` (N ` A)

proof

let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in N ` A or x in N ` (N ` A) )
let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in N ` (N ` A) or x in N ` A )

assume x in N ` (N ` A) ; :: thesis: x in N ` A

then consider y being Element of G such that

A1: ( y = x & y * N c= N ` A ) ;

y in y * N by GROUP_2:108;

hence x in N ` A by A1; :: thesis: verum

end;assume x in N ` (N ` A) ; :: thesis: x in N ` A

then consider y being Element of G such that

A1: ( y = x & y * N c= N ` A ) ;

y in y * N by GROUP_2:108;

hence x in N ` A by A1; :: thesis: verum

assume A2: x in N ` A ; :: thesis: x in N ` (N ` A)

then reconsider x9 = x as Element of G ;

A3: x9 * N c= A by A2, Th12;

x9 * N c= N ` A

proof

hence
x in N ` (N ` A)
; :: thesis: verum
let y be object ; :: according to TARSKI:def 3 :: thesis: ( not y in x9 * N or y in N ` A )

assume A4: y in x9 * N ; :: thesis: y in N ` A

then reconsider y9 = y as Element of G ;

x9 * N = y9 * N by A4, Th2;

hence y in N ` A by A3; :: thesis: verum

end;assume A4: y in x9 * N ; :: thesis: y in N ` A

then reconsider y9 = y as Element of G ;

x9 * N = y9 * N by A4, Th2;

hence y in N ` A by A3; :: thesis: verum