 reserve G, A for Group;
 reserve phi for Homomorphism of A,AutGroup(G);
 reserve G, A for Group;
 reserve phi for Homomorphism of A,AutGroup(G);
reserve G1,G2 for Group;

theorem Th106:
  for n being non zero Nat
  for g1 being Element of INT.Group n
  for a2 being Element of INT.Group 2 st a2 = 1
  for x,y being Element of Dihedral_group n
  st x = <*g1,1_(INT.Group 2)*> & y = <*(1_(INT.Group n)),a2*>
  for i being Nat
  holds y*(x |^ i) = (x |^ (n - i))*y
proof
  let n be non zero Nat;
  let g1 be Element of INT.Group n;
  let a2 be Element of INT.Group 2;
  assume A1: a2 = 1;
  let x,y be Element of Dihedral_group n;
  assume A2: x = <*g1,1_(INT.Group 2)*>;
  assume A3: y = <*(1_(INT.Group n)),a2*>;
  defpred P[Nat] means y*(x |^ $1) = (x |^ (n - $1))*y;
  A4: P[0]
  proof
    thus y * (x |^ 0) = y * 1_(Dihedral_group n) by GROUP_1:25
                     .= y by GROUP_1:def 4
                     .= (1_(Dihedral_group n))*y by GROUP_1:def 4
                     .= (x |^ n)*y by A2,Th101
                     .= (x |^ (n - 0))*y;
  end;
  A5: for k being Nat st P[k] holds P[k + 1]
  proof
    let k be Nat;
    assume B1: P[k];
    y * (x |^ (k + 1)) = y * ((x |^ k) * x) by GROUP_1:34
                      .= (y * (x |^ k)) * x by GROUP_1:def 3
                      .= ((x |^ (n - k)) * y) * x by B1
                      .= (x |^ (n - k)) * (y * x) by GROUP_1:def 3
                      .= (x |^ (n - k)) * ((x |^ (- 1)) * y) by A1,A2,A3,Th105
                      .= ((x |^ (n - k)) * (x |^ (- 1))) * y by GROUP_1:def 3
                      .= (x |^ ((n - k) + (- 1))) * y by GROUP_1:33
                      .= (x |^ (n - (k + 1))) * y;
    hence P[k + 1];
  end;
  for k being Nat holds P[k] from NAT_1:sch 2(A4, A5);
  hence thesis;
end;
