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\(x\(y\(y\x)))) <= (x\(x\y))\(y\x)
proof
  thus X is BCK-positive-implicative BCK-algebra implies for x,y being Element
  of X holds (x\(x\(y\(y\x)))) <= (x\(x\y))\(y\x)
  proof
    assume
A1: X is BCK-positive-implicative BCK-algebra;
    let x,y be Element of X;
    (x\(x\(y\(y\x)))) \((x\(x\y))\(y\x)) = ((x\(x\(y\(y\x)))))\((x\(x\(y\(
    y\x))))) by A1,Th29
      .= 0.X by BCIALG_1:def 5;
    hence thesis;
  end;
  assume
A2: for x,y being Element of X holds (x\(x\(y\(y\x)))) <= (x\(x\y))\(y\x );
  for x,y being Element of X holds (x\(x\y))\(y\x) = x\(x\(y\(y\x)))
  proof
    let x,y be Element of X;
    (x\(x\(y\(y\x)))) <= (x\(x\y))\(y\x) by A2;
    then
A3: (x\(x\(y\(y\x))))\((x\(x\y))\(y\x)) = 0.X;
    ((x\(x\y))\(y\x))\(x\(x\(y\(y\x))))=0.X by BCIALG_1:10;
    hence thesis by A3,BCIALG_1:def 7;
  end;
  hence thesis by Th29;
end;
