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

theorem
  X is BCK-positive-implicative BCK-algebra iff for x,y being Element of
  X holds x\y = (x\y)\(x\(x\y))
proof
  thus X is BCK-positive-implicative BCK-algebra implies for x,y being Element
  of X holds x\y = (x\y)\(x\(x\y))
  proof
    assume
A1: X is BCK-positive-implicative BCK-algebra;
    let x,y be Element of X;
A2: ((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\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;
    hence thesis by A1,A2,Th29;
  end;
  assume
A3: for x,y being Element of X holds x\y = (x\y)\(x\(x\y));
  for x,y being Element of X holds x\y = (x\y)\y
  proof
    let x,y be Element of X;
    (x\y)\(x\(x\y)) = (x\(x\(x\y)))\y by BCIALG_1:7
      .= (x\y)\y by BCIALG_1:8;
    hence thesis by A3;
  end;
  hence thesis by Th28;
end;
