reserve X for BCK-algebra;
reserve x,y for Element of X;
reserve IT for non empty Subset of X;

theorem
  X is BCK-positive-implicative BCK-algebra iff for x,y being Element of
  X holds x\y <= (x\y)\y
proof
  thus X is BCK-positive-implicative BCK-algebra implies for x,y being Element
  of X holds x\y <= (x\y)\y
  proof
    assume
A1: X is BCK-positive-implicative BCK-algebra;
    let x,y be Element of X;
    (x\y)\((x\y)\y) = ((x\y)\y)\((x\y)\y) by A1,Th28
      .= 0.X by BCIALG_1:def 5;
    hence thesis;
  end;
  assume
A2: for x,y being Element of X holds x\y <= (x\y)\y;
  for x,y being Element of X holds x\y = (x\y)\y
  proof
    let x,y be Element of X;
    x\y <= (x\y)\y by A2;
    then
A3: (x\y)\((x\y)\y) = 0.X;
    ((x\y)\y)\(x\y) = ((x\y)\(x\y))\y by BCIALG_1:7
      .= y` by BCIALG_1:def 5
      .= 0.X by BCIALG_1:def 8;
    hence thesis by A3,BCIALG_1:def 7;
  end;
  hence thesis by Th28;
end;
