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

theorem Th29:
  X is BCK-positive-implicative BCK-algebra iff for x,y being
  Element of X holds (x\(x\y))\(y\x) = x\(x\(y\(y\x)))
proof
  thus X is BCK-positive-implicative BCK-algebra implies for x,y being Element
  of X holds (x\(x\y))\(y\x) = x\(x\(y\(y\x)))
  proof
    assume
A1: X is BCK-positive-implicative BCK-algebra;
    for x,y being Element of X holds (x\(x\y))\(y\x) = x\(x\(y\(y\x)))
    proof
      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;
      then (x\(x\y)) <= y;
      then
A2:   (x\(x\y))\(y\(y\x)) <= y\(y\(y\x)) by BCIALG_1:5;
      (x\(x\y))\(x\(x\(y\(y\x)))) = (x\(x\(x\(y\(y\x)))))\(x\y) by BCIALG_1:7
        .= (x\(y\(y\x)))\(x\y) by BCIALG_1:8
        .= (x\(x\y))\(y\(y\x)) by BCIALG_1:7;
      then (x\(x\y))\(x\(x\(y\(y\x)))) <= y\x by A2,BCIALG_1:8;
      then ((x\(x\y))\(x\(x\(y\(y\x)))))\(y\x) = 0.X;
      then
A3:   ((x\(x\y))\(y\x))\(x\(x\(y\(y\x)))) =0.X by BCIALG_1:7;
      (y\(y\x))\x = (y\x)\(y\x) by BCIALG_1:7
        .= 0.X by BCIALG_1:def 5;
      then (y\(y\x)) <= x;
      then (y\(y\x))\(y\x) <= x\(y\x) by BCIALG_1:5;
      then (y\(y\x)) <= x\(y\x) by A1,Th28;
      then
A4:   (y\(y\x))\(x\(y\x)) = 0.X;
      (x\(x\(y\(y\x))))\(y\(y\x)) = (x\(y\(y\x)))\(x\(y\(y\x))) by BCIALG_1:7
        .= 0.X by BCIALG_1:def 5;
      then (x\(x\(y\(y\x))))\(x\(y\x)) = 0.X by A4,BCIALG_1:3;
      then (x\(x\(y\(y\x)))) <= (x\(y\x));
      then (x\(x\(y\(y\x))))\(x\(y\(y\x))) <= (x\(y\x))\(x\(y\(y\x))) by
BCIALG_1:5;
      then (x\(x\(y\(y\x))))\(x\(y\(y\x))) <= (x\(x\(y\(y\x))))\(y\x) by
BCIALG_1:7;
      then (x\(x\(y\(y\x)))) <= (x\(x\(y\(y\x))))\(y\x) by A1,Th28;
      then
A5:   (x\(x\(y\(y\x))))\((x\(x\(y\(y\x))))\(y\x)) = 0.X;
      (y\(y\x))\y = (y\y)\(y\x) by BCIALG_1:7
        .= (y\x)` by BCIALG_1:def 5
        .= 0.X by BCIALG_1:def 8;
      then (y\(y\x)) <= y;
      then x\y <= x\(y\(y\x)) by BCIALG_1:5;
      then x\(x\(y\(y\x))) <= x\(x\y) by BCIALG_1:5;
      then (x\(x\(y\(y\x))))\(y\x) <= x\(x\y)\(y\x) by BCIALG_1:5;
      then ((x\(x\(y\(y\x))))\(y\x))\(x\(x\y)\(y\x)) = 0.X;
      then (x\(x\(y\(y\x))))\(x\(x\y)\(y\x)) = 0.X by A5,BCIALG_1:3;
      hence thesis by A3,BCIALG_1:def 7;
    end;
    hence thesis;
  end;
  assume
A6: for x,y being Element of X holds (x\(x\y))\(y\x) = x\(x\(y\(y\x)));
  for x,y being Element of X holds x\y = (x\y)\y
  proof
    let x,y be Element of X;
A7: (x\y)\((x\y)\(x\(x\(x\y)))) = (x\y)\((x\y)\(x\y)) by BCIALG_1:8
      .= (x\y)\0.X by BCIALG_1:def 5
      .= x\y by BCIALG_1:2;
    ((x\y)\((x\y)\x))\(x\(x\y))=((x\y)\((x\x)\y))\(x\(x\y)) by BCIALG_1:7
      .=((x\y)\y`)\(x\(x\y)) by BCIALG_1:def 5
      .=((x\y)\0.X)\(x\(x\y)) by BCIALG_1:def 8
      .=(x\y)\(x\(x\y)) by BCIALG_1:2
      .=(x\(x\(x\y)))\y by BCIALG_1:7
      .=(x\y)\y by BCIALG_1:8;
    hence thesis by A6,A7;
  end;
  hence thesis by Th28;
end;
