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 Th47:
  for X being BCK-Algebra_with_Condition(S) holds ( X is
  positive-implicative iff for x,y being Element of X holds x*y = x*(y\x) )
proof
  let X be BCK-Algebra_with_Condition(S);
A1: X is positive-implicative implies for x,y being Element of X holds x*y =
  x*(y\x)
  proof
    assume
A2: X is positive-implicative;
    let x,y be Element of X;
    (y\x)\y = (y\y)\x by BCIALG_1:7
      .= x` by BCIALG_1:def 5
      .= 0.X by BCIALG_1:def 8;
    then y\x <= y;
    then x*(y\x) <= x*y by Th7;
    then
A3: (x*(y\x))\(x*y) = 0.X;
    (x*y)\x = (x\x)*(y\x) by A2,Th46
      .= 0.X*(y\x) by BCIALG_1:def 5
      .= y\x by Th8;
    then ((x*y)\x)\(y\x) = 0.X by BCIALG_1:def 5;
    then (x*y)\(x*(y\x)) = 0.X by Th11;
    hence thesis by A3,BCIALG_1:def 7;
  end;
  (for x,y being Element of X holds x*y = x*(y\x)) implies X is
  positive-implicative
  proof
    assume
A4: for x,y being Element of X holds x*y = x*(y\x);
    for x,y being Element of X holds (x\y)\y = x\y
    proof
      let x,y be Element of X;
      (y*y) = y*(y\y) by A4
        .= y*0.X by BCIALG_1:def 5
        .= y by Th8;
      hence thesis by Th11;
    end;
    hence thesis;
  end;
  hence thesis by A1;
end;
