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

theorem Th16:
  for X being BCI-algebra holds (X is BCI-commutative iff for x,y
  being Element of X holds x\(x\y) = y\(y\(x\(x\y))) )
proof
  let X be BCI-algebra;
A1: (for x,y being Element of X holds x\(x\y) = y\(y\(x\(x\y))) ) implies X
  is BCI-commutative
  proof
    assume
A2: for x,y being Element of X holds x\(x\y) = y\(y\(x\(x\y)));
    let x,y be Element of X;
    assume x\y=0.X;
    then x\(x\y) = x by BCIALG_1:2;
    hence thesis by A2;
  end;
  X is BCI-commutative implies for x,y being Element of X holds x\(x\y) =
  y\(y\(x\(x\y)))
  proof
    assume
A3: X is BCI-commutative;
    let x,y be Element of X;
    (x\(x\y))\y = (x\y)\(x\y) by BCIALG_1:7
      .= 0.X by BCIALG_1:def 5;
    hence thesis by A3;
  end;
  hence thesis by A1;
end;
