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