reserve i for Element of NAT;

theorem Th13:
  for G,H being strict Group, h being Homomorphism of G,H for A
being strict Subgroup of G for a being Element of G holds h.a* h.:A=h.:(a*A) &
  h.:A * h.a=h.:(A*a)
proof
  let G,H be strict Group;
  let h be Homomorphism of G,H;
  let A be strict Subgroup of G;
  let a be Element of G;
  now
    let x be object;
    assume x in h.:(a *A);
    then consider y being object such that
A1: y in the carrier of G and
A2: y in a*A and
A3: x = h.y by FUNCT_2:64;
    reconsider y as Element of G by A1;
    consider b being Element of G such that
A4: y=a* b and
A5: b in A by A2,GROUP_2:103;
    b in the carrier of A by A5,STRUCT_0:def 5;
    then h.b in h.:(the carrier of A) by FUNCT_2:35;
    then h.b in the carrier of h.: A by Th8;
    then
A6: h.b in h.:A by STRUCT_0:def 5;
    x=h.a *h.b by A3,A4,GROUP_6:def 6;
    hence x in h.a * h.:A by A6,GROUP_2:103;
  end;
  then
A7: h.:(a * A) c= h.a *h.:A;
  now
    let x be object;
    assume x in h.a *h.:A;
    then consider b1 being Element of H such that
A8: x = h.a * b1 and
A9: b1 in h.:A by GROUP_2:103;
    consider b being Element of A such that
A10: b1=(h|A).b by A9,GROUP_6:45;
A11: b in A by STRUCT_0:def 5;
    reconsider b as Element of G by GROUP_2:42;
    b1=h.b by A10,FUNCT_1:49;
    then
A12: x=h.(a*b) by A8,GROUP_6:def 6;
    a* b in a*A by A11,GROUP_2:103;
    hence x in h.:(a*A) by A12,FUNCT_2:35;
  end;
  then h.a *h.:A c= h.:(a * A);
  hence h.a *h.:A = h.:(a * A) by A7,XBOOLE_0:def 10;
  now
    let x be object;
    assume x in h.:(A* a);
    then consider y being object such that
A13: y in the carrier of G and
A14: y in A* a and
A15: x = h.y by FUNCT_2:64;
    reconsider y as Element of G by A13;
    consider b being Element of G such that
A16: y=b* a and
A17: b in A by A14,GROUP_2:104;
    b in the carrier of A by A17,STRUCT_0:def 5;
    then h.b in h.:(the carrier of A) by FUNCT_2:35;
    then h.b in the carrier of h.: A by Th8;
    then
A18: h.b in h.:A by STRUCT_0:def 5;
    x=h.b *h.a by A15,A16,GROUP_6:def 6;
    hence x in h.:A*h.a by A18,GROUP_2:104;
  end;
  then
A19: h.:(A*a) c= h.:A*h.a;
  now
    let x be object;
    assume x in h.:A*h.a;
    then consider b1 being Element of H such that
A20: x = b1*h.a and
A21: b1 in h.:A by GROUP_2:104;
    consider b being Element of A such that
A22: b1=(h|A).b by A21,GROUP_6:45;
A23: b in A by STRUCT_0:def 5;
    reconsider b as Element of G by GROUP_2:42;
    b1=h.b by A22,FUNCT_1:49;
    then
A24: x=h.(b*a) by A20,GROUP_6:def 6;
    b* a in A*a by A23,GROUP_2:104;
    hence x in h.:(A*a) by A24,FUNCT_2:35;
  end;
  then h.:A*h.a c= h.:(A* a);
  hence thesis by A19,XBOOLE_0:def 10;
end;
