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

theorem Th34:
  X is BCK-implicative BCK-algebra iff X is commutative
  BCK-algebra & X is BCK-positive-implicative BCK-algebra
proof
  thus X is BCK-implicative BCK-algebra implies X is commutative BCK-algebra &
  X is BCK-positive-implicative BCK-algebra
  proof
    assume
A1: X is BCK-implicative BCK-algebra;
A2: for x,y being Element of X holds x\(x\y) <= 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 (x\(x\y))\(y\x) <= y\(y\x) by BCIALG_1:5;
      then (x\(y\x))\(x\y) <= y\(y\x) by BCIALG_1:7;
      hence thesis by A1,Def12;
    end;
    for x,y being Element of X holds x\y = (x\y)\y
    proof
      let x,y be Element of X;
      (x\y)\(y\(x\y))=(x\y) by A1,Def12;
      hence thesis by A1,Def12;
    end;
    hence thesis by A2,Th1,Th28;
  end;
  assume that
A3: X is commutative BCK-algebra and
A4: X is BCK-positive-implicative BCK-algebra;
  for x,y being Element of X holds x\(y\x)=x
  proof
    let x,y be Element of X;
    x\(x\(x\(y\x))) = x\(y\x) by BCIALG_1:8;
    then
A5: x\(y\x) = x\((y\x)\((y\x)\x)) by A3,Def1;
    (y\x)\((y\x)\x) = (y\x)\(y\x) by A4,Th28
      .= 0.X by BCIALG_1:def 5;
    hence thesis by A5,BCIALG_1:2;
  end;
  hence thesis by Def12;
end;
