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

theorem
  for X being BCI-algebra holds (X is BCI-commutative iff for x,y,z
  being Element of X st x<=y & x<=z holds x<=y\(y\z) )
proof
  let X be BCI-algebra;
  thus X is BCI-commutative implies for x,y,z being Element of X st x<=y & x<=
  z holds x<=y\(y\z)
  proof
    assume
A1: X is BCI-commutative;
    for x,y,z being Element of X st x<=y & x<=z holds x<=y\(y\z)
    proof
      let x,y,z be Element of X;
      assume that
A2:   x<=y and
A3:   x<=z;
A4:   x\z = 0.X by A3;
      x\y = 0.X by A2;
      then
A5:   x = y\(y\x) by A1;
      then x\(y\(y\z)) = (y\(y\(y\z)))\(y\x) by BCIALG_1:7
        .= (y\z)\(y\x) by BCIALG_1:8
        .= 0.X by A4,A5,BCIALG_1:7;
      hence thesis;
    end;
    hence thesis;
  end;
  assume
A6: for x,y,z being Element of X st x<=y & x<=z holds x<=y\(y\z);
  for x,y being Element of X st x\y=0.X holds x = y\(y\x)
  proof
    let x,y be Element of X;
    x\x =0.X by BCIALG_1:def 5;
    then
A7: x<=x;
    assume x\y=0.X;
    then x<=y;
    then x<=y\(y\x) by A6,A7;
    then
A8: x\(y\(y\x)) = 0.X;
    (y\(y\x))\x = (y\x)\(y\x) by BCIALG_1:7
      .= 0.X by BCIALG_1:def 5;
    hence thesis by A8,BCIALG_1:def 7;
  end;
  hence thesis;
end;
