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

theorem
  X is BCK-implicative BCK-algebra iff for x,y being Element of X holds
  (x\(x\y)) = (y\(y\x))\(x\y)
proof
  thus X is BCK-implicative BCK-algebra implies for x,y being Element of X
  holds (x\(x\y)) = (y\(y\x))\(x\y)
  proof
    assume
A1: X is BCK-implicative BCK-algebra;
    then
A2: X is commutative BCK-algebra by Th34;
    for x,y being Element of X holds x\(x\y) = (y\(y\x))\(x\y)
    proof
      let x,y be Element of X;
      (x\(x\y))\(x\y) = y\(y\x) by A1,Th35;
      then (y\(y\x))\(x\y) = y\(y\x) by A2,Def1;
      hence thesis by A2,Def1;
    end;
    hence thesis;
  end;
  assume
A3: for x,y being Element of X holds (x\(x\y)) = (y\(y\x))\(x\y);
  for x,y being Element of X st x<= y holds x= y\(y\x)
  proof
    let x,y be Element of X;
    assume x<=y;
    then x\y = 0.X;
    then (x\0.X) = (y\(y\x))\0.X by A3;
    then (x\0.X) = y\(y\x) by BCIALG_1:2;
    hence thesis by BCIALG_1:2;
  end;
  then
A4: X is commutative BCK-algebra by Th5;
  for x,y being Element of X holds x\y = (x\y)\y
  proof
    let x,y be Element of X;
A5: (x\y)\x = (x\x)\y by BCIALG_1:7
      .= y` by BCIALG_1:def 5
      .= 0.X by BCIALG_1:def 8;
    (x\(x\(x\y))) = ((x\y)\((x\y)\x))\(x\(x\y)) by A3;
    then (x\(x\(x\y))) = (x\y)\(x\(x\y)) by A5,BCIALG_1:2
      .= (x\(x\(x\y)))\y by BCIALG_1:7
      .= (x\y)\y by BCIALG_1:8;
    hence thesis by BCIALG_1:8;
  end;
  then X is BCK-positive-implicative BCK-algebra by Th28;
  hence thesis by A4,Th34;
end;
