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,z being Element
  of X holds (x\z)\(y\z) <= (x\y)\z
proof
  thus X is BCK-positive-implicative BCK-algebra implies for x,y,z being
  Element of X holds (x\z)\(y\z) <= (x\y)\z
  proof
    assume
A1: X is BCK-positive-implicative BCK-algebra;
    let x,y,z be Element of X;
    ((x\z)\(y\z))\((x\y)\z) = ((x\z)\(y\z))\((x\z)\(y\z)) by A1,Def11
      .= 0.X by BCIALG_1:def 5;
    hence thesis;
  end;
  assume
A2: for x,y,z being Element of X holds (x\z)\(y\z) <= (x\y)\z;
  for x,y,z being Element of X holds (x\y)\z=(x\z)\(y\z)
  proof
    let x,y,z be Element of X;
    (y\z)\y = (y\y)\z by BCIALG_1:7
      .= z` by BCIALG_1:def 5
      .= 0.X by BCIALG_1:def 8;
    then y\z <= y;
    then x\y <= x\(y\z) by BCIALG_1:5;
    then ((x\y)\z) <= ((x\(y\z))\z) by BCIALG_1:5;
    then ((x\y)\z)\((x\(y\z))\z) = 0.X;
    then
A3: ((x\y)\z)\((x\z)\(y\z)) = 0.X by BCIALG_1:7;
    (x\z)\(y\z) <= (x\y)\z by A2;
    then ((x\z)\(y\z))\((x\y)\z) = 0.X;
    hence thesis by A3,BCIALG_1:def 7;
  end;
  hence thesis by Def11;
end;
