reserve X for BCI-algebra;
reserve I for Ideal of X;
reserve a,x,y,z,u for Element of X;
reserve f,f9,g for sequence of  the carrier of X;
reserve j,i,k,n,m for Nat;

theorem
  ((x,z) to_power n)\((y,z) to_power n) <= x\y
proof
  defpred P[set] means for m being Nat holds m=$1 & m<=n implies ((
  x,z) to_power m)\((y,z) to_power m) <= x\y;
A1: for k st P[k] holds P[k+1]
  proof
    let k;
    assume
A2: for m being Nat holds m=k & m<= n implies((x,z)
    to_power m)\((y,z) to_power m) <= x\y;
    let m be Nat;
    assume that
A3: m=k+1 and
A4: m<=n;
    k<=n by A3,A4,NAT_1:13;
    then ((x,z) to_power k)\((y,z) to_power k) <= x\y by A2;
    then (((x,z) to_power k)\((y,z) to_power k))\(x\y)=0.X;
    then (((x,z) to_power k)\(x\y))\((y,z) to_power k)=0.X by BCIALG_1:7;
    then (((x,z) to_power k)\(x\y))\z\(((y,z) to_power k)\z)=0.X by BCIALG_1:4;
    then (((x,z) to_power k)\z)\(x\y)\(((y,z) to_power k)\z)=0.X by BCIALG_1:7;
    then ((x,z) to_power (k+1))\(x\y)\(((y,z) to_power k)\z)=0.X by Th4;
    then ((x,z) to_power (k+1))\(x\y)\((y,z) to_power (k+1))=0.X by Th4;
    then ((x,z) to_power (k+1))\((y,z) to_power (k+1))\(x\y)=0.X by BCIALG_1:7;
    hence thesis by A3;
  end;
  (x\y)\(x\y)=0.X by BCIALG_1:def 5;
  then x\y <= x\y;
  then ((x,z) to_power 0)\y <= x\y by Th1;
  then
A5: P[0] by Th1;
  for n holds P[n] from NAT_1:sch 2(A5,A1);
  hence thesis;
end;
