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

theorem Th40:
  X is BCK-implicative BCK-algebra iff for x,y,z being Element of
  X holds (x\z)\(x\y) = (y\z)\((y\x)\z)
proof
  thus X is BCK-implicative BCK-algebra implies for x,y,z being Element of X
  holds (x\z)\(x\y) = (y\z)\((y\x)\z)
  proof
    assume
A1: X is BCK-implicative BCK-algebra;
    then
A2: X is commutative BCK-algebra by Th34;
    let x,y,z be Element of X;
A3: (x\z)\(x\y) = (x\(x\y))\z by BCIALG_1:7
      .= (y\(y\x))\z by A2,Def1;
    X is BCK-positive-implicative BCK-algebra by A1,Th34;
    hence thesis by A3,Def11;
  end;
  assume
A4: for x,y,z being Element of X holds (x\z)\(x\y)=(y\z)\((y\x)\z);
A5: for x,y being Element of X holds x\(x\y)= y\(y\x)
  proof
    let x,y be Element of X;
    (x\0.X)\(x\y) = (y\0.X)\((y\x)\0.X) by A4;
    then (x\0.X)\(x\y) = y\((y\x)\0.X) by BCIALG_1:2;
    then (x\0.X)\(x\y) = y\(y\x) by BCIALG_1:2;
    hence thesis by BCIALG_1:2;
  end;
  for x,y being Element of X holds (y\(y\x))\(y\x) = x\(x\y)
  proof
    let x,y be Element of X;
    (x\(y\x))\(x\y) = (y\(y\x))\((y\x)\(y\x)) by A4;
    then (x\(y\x))\(x\y) = (y\(y\x))\0.X by BCIALG_1:def 5;
    then (x\(y\x))\(x\y) = y\(y\x) by BCIALG_1:2;
    then (x\(x\y))\(y\x) = y\(y\x) by BCIALG_1:7;
    then (y\(y\x))\(y\x) = y\(y\x) by A5;
    hence thesis by A5;
  end;
  then for x,y being Element of X holds (x\(x\y))\(x\y) = y\(y\x);
  hence thesis by Th35;
end;
