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

theorem Th17:
  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)) = 0.X\(x\y) )
proof
  let X be BCI-algebra;
A1: (for x,y being Element of X holds (x\(x\y))\(y\(y\x)) = 0.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)) = 0.X\(x\y);
    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;
A3:   (y\(y\x))\x = (y\x)\(y\x) by BCIALG_1:7
        .= 0.X by BCIALG_1:def 5;
      assume
A4:   x\y=0.X;
      then 0.X = 0.X\(x\y) by BCIALG_1:2
        .= (x\0.X)\(y\(y\x)) by A2,A4
        .= x\(y\(y\x)) by BCIALG_1:2;
      hence thesis by A3,BCIALG_1:def 7;
    end;
    hence thesis;
  end;
  X is BCI-commutative implies for x,y being Element of X holds (x\(x\y))\
  (y\(y\x)) = 0.X\(x\y)
  proof
    assume
A5: X is BCI-commutative;
    let x,y be Element of X;
    (x\(x\y))\(y\(y\x)) = (y\(y\(x\(x\y))))\(y\(y\x)) by A5,Th16
      .= (y\(y\(y\x)))\(y\(x\(x\y))) by BCIALG_1:7
      .= (y\x)\(y\(x\(x\y))) by BCIALG_1:8
      .= (y\(y\(x\(x\y))))\x by BCIALG_1:7
      .= (x\(x\y))\x by A5,Th16
      .= (x\x)\(x\y) by BCIALG_1:7;
    hence thesis by BCIALG_1:def 5;
  end;
  hence thesis by A1;
end;
