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 Th50:
  for X being BCK-Algebra_with_Condition(S) holds (X is
  implicative iff X is commutative & X is positive-implicative )
proof
  let X be BCK-Algebra_with_Condition(S);
  thus X is implicative implies X is commutative & X is positive-implicative
  proof
    assume
A1: X is implicative;
A2: for x,y being Element of X holds x\(x\y) <= y\(y\x)
    proof
      let x,y be Element of X;
      (x\(x\y))\y = (x\y)\(x\y) by BCIALG_1:7
        .= 0.X by BCIALG_1:def 5;
      then (x\(x\y)) <= y;
      then (x\(x\y))\(y\x) <= y\(y\x) by BCIALG_1:5;
      then (x\(y\x))\(x\y) <= y\(y\x) by BCIALG_1:7;
      hence thesis by A1;
    end;
A3: for x,y being Element of X holds x\(x\y) = y\(y\x)
    proof
      let x,y be Element of X;
      y\(y\x) <= x\(x\y) by A2;
      then
A4:   (y\(y\x))\(x\(x\y)) = 0.X;
      x\(x\y) <= y\(y\x) by A2;
      then (x\(x\y))\(y\(y\x)) = 0.X;
      hence thesis by A4,BCIALG_1:def 7;
    end;
    for x,y being Element of X holds x\y = (x\y)\y
    proof
      let x,y be Element of X;
      (x\y)\(y\(x\y))=(x\y) by A1;
      hence thesis by A1;
    end;
    hence thesis by A3;
  end;
  assume that
A5: X is commutative and
A6: X is positive-implicative;
  for x,y being Element of X holds x\(y\x)=x
  proof
    let x,y be Element of X;
    x\(x\(x\(y\x))) = x\(y\x) by BCIALG_1:8;
    then
A7: x\(y\x) = x\((y\x)\((y\x)\x)) by A5;
    (y\x)\((y\x)\x) = (y\x)\(y\x) by A6
      .= 0.X by BCIALG_1:def 5;
    hence thesis by A7,BCIALG_1:2;
  end;
  hence thesis;
end;
