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

theorem Th35:
  X is BCK-implicative BCK-algebra iff for x,y being Element of X
  holds (x\(x\y))\(x\y) = (y\(y\x))
proof
  thus X is BCK-implicative BCK-algebra implies for x,y being Element of X
  holds (x\(x\y))\(x\y) = (y\(y\x))
  proof
    assume
A1: X is BCK-implicative BCK-algebra;
    let x,y be Element of X;
    X is commutative BCK-algebra by A1,Th34;
    then (x\(x\y))\(x\y) = (y\(y\x))\(x\y) by Def1
      .= (y\(x\y))\(y\x) by BCIALG_1:7;
    hence thesis by A1,Def12;
  end;
  assume
A2: for x,y being Element of X holds (x\(x\y))\(x\y)=y\(y\x);
  for x,y being Element of X holds (x\y)\y = x\y
  proof
    let x,y be Element of X;
A3: (x\(x\(x\y)))\(x\(x\y)) = (x\y)\(x\(x\y)) by BCIALG_1:8
      .= (x\(x\(x\y)))\y by BCIALG_1:7
      .= (x\y)\y by BCIALG_1:8;
A4: (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\(x\y)) = ((x\y)\((x\y)\x)) by A2;
    hence thesis by A3,A4,BCIALG_1:2;
  end;
  then
A5: X is BCK-positive-implicative BCK-algebra by Th28;
  for x,y being Element of X holds ( x<= y implies x= y\(y\x) )
  proof
    let x,y be Element of X;
    assume x<=y;
    then
A6: x\y = 0.X;
    then (y\(y\x)) = (x\(x\y))\0.X by A2
      .= (x\0.X) by A6,BCIALG_1:2;
    hence thesis by BCIALG_1:2;
  end;
  then X is commutative BCK-algebra by Th5;
  hence thesis by A5,Th34;
end;
