reserve X for non empty BCIStr_1;
reserve d for Element of X;
reserve n,m,k for Nat;
reserve f for sequence of  the carrier of X;

theorem
  for X being BCK-Algebra_with_Condition(S) holds ( X is implicative iff
  for x,y,z being Element of X holds x\(y\z) = ((x\y)\z)*(z\(z\x)) )
proof
  let X be BCK-Algebra_with_Condition(S);
A1: X is implicative implies for x,y,z being Element of X holds x\(y\z) = ((
  x\y)\z)*(z\(z\x))
  proof
    assume
A2: X is implicative;
    then
A3: X is positive-implicative by Th50;
    let x,y,z be Element of X;
A4: X is commutative by A2,Th50;
    x = (x\z)*(x\(x\z)) by A3,Th48;
    then
A5: x\(y\z) = ((x\z)*(z\(z\x)))\(y\z) by A4;
    (y\z)\y = (y\y)\z by BCIALG_1:7
      .= z` by BCIALG_1:def 5
      .= 0.X by BCIALG_1:def 8;
    then y\z <= y;
    then (x\z)\y <= (x\z)\(y\z) by BCIALG_1:5;
    then ((x\z)\y)\((x\z)\(y\z)) = 0.X;
    then
A6: ((x\y)\z)\((x\z)\(y\z)) = 0.X by BCIALG_1:7;
    ((x\z)\(y\z))\((x\y)\z) = (((x\z)\z)\(y\z))\((x\y)\z) by A3
      .= (((x\z)\z)\(y\z))\((x\z)\y) by BCIALG_1:7
      .= 0.X by BCIALG_1:def 3;
    then
A7: (x\y)\z=(x\z)\(y\z) by A6,BCIALG_1:def 7;
    (z\(z\x))\(y\z) = (z\(y\z))\(z\x) by BCIALG_1:7
      .= z\(z\x) by A2;
    hence thesis by A3,A5,A7,Th46;
  end;
  (for x,y,z being Element of X holds x\(y\z) = ((x\y)\z)*(z\(z\x)))
  implies X is implicative
  proof
    assume
A8: for x,y,z being Element of X holds x\(y\z) = ((x\y)\z)*(z\(z\x));
    for x,y being Element of X holds x\(y\x) = x
    proof
      let x,y be Element of X;
      x\(y\x) = ((x\y)\x)*(x\(x\x)) by A8
        .= ((x\y)\x)*(x\0.X) by BCIALG_1:def 5
        .= ((x\y)\x)*x by BCIALG_1:2
        .= ((x\x)\y)*x by BCIALG_1:7
        .= (y`)*x by BCIALG_1:def 5
        .= 0.X*x by BCIALG_1:def 8;
      hence thesis by Th8;
    end;
    hence thesis;
  end;
  hence thesis by A1;
end;
