reserve X for BCI-algebra;
reserve x,y,z for Element of X;
reserve i,j,k,l,m,n for Nat;
reserve f,g for sequence of the carrier of X;
reserve B,P for non empty Subset of X;

theorem Th49:
  X is BCK-positive-implicative BCK-algebra iff X is BCK-algebra of 0,1,0,1
proof
  thus X is BCK-positive-implicative BCK-algebra implies X is BCK-algebra of 0
  ,1,0,1
  proof
    assume
A1: X is BCK-positive-implicative BCK-algebra;
    for x,y being Element of X holds Polynom (0,1,x,y) = Polynom (0,1,y,x)
    proof
      let x,y be Element of X;
      (x\y)\(x\y) = 0.X by BCIALG_1:def 5;
      then (x\(x\y))\y = 0.X by BCIALG_1:7;
      then
A2:   x\(x\y) <= y;
      x\(x\y) = (x\(x\y))\(x\y) by A1,BCIALG_3:28;
      then x\(x\y) <= y\(x\y) by A2,BCIALG_1:5;
      then (x\(x\y))\(y\x) <= (y\(x\y))\(y\x) by BCIALG_1:5;
      then
A3:   (x\(x\y))\(y\x) <= (y\(y\x))\(x\y) by BCIALG_1:7;
      (y\x)\(y\x) = 0.X by BCIALG_1:def 5;
      then (y\(y\x))\x = 0.X by BCIALG_1:7;
      then y\(y\x) <= x;
      then
A4:   (y\(y\x))\(y\x) <= x\(y\x) by BCIALG_1:5;
      y\(y\x) = (y\(y\x))\(y\x) by A1,BCIALG_3:28;
      then (y\(y\x))\(x\y) <= (x\(y\x))\(x\y) by A4,BCIALG_1:5;
      then (y\(y\x))\(x\y) <= (x\(x\y))\(y\x) by BCIALG_1:7;
      then
A5:   (x\(x\y))\(y\x) = (y\(y\x))\(x\y) by A3,Th2;
      ((x,(x\y)) to_power 1,(y\x)) to_power 1 = (x,(x\y)) to_power 1 \ (y
      \x) by BCIALG_2:2
        .= (x\(x\y))\(y\x) by BCIALG_2:2
        .= ((y\(y\x)),(x\y)) to_power 1 by A5,BCIALG_2:2
        .= ((y,(y\x)) to_power 1,(x\y)) to_power 1 by BCIALG_2:2;
      hence thesis;
    end;
    hence thesis by A1,Def3;
  end;
  assume
A6: X is BCK-algebra of 0,1,0,1;
  for x,y being Element of X holds x\y = (x\y)\y
  proof
    let x,y be Element of X;
A7: Polynom (0,1,x,(x\y)) = Polynom (0,1,(x\y),x) by A6,Def3;
A8: (x\y)\x = (x\x)\y by BCIALG_1:7
      .= y` by BCIALG_1:def 5
      .= 0.X by A6,BCIALG_1:def 8;
    then
A9: ((x\y)\((x\y)\x))\(x\(x\y)) = (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;
A10: (x\(x\(x\y)))\((x\y)\x) = (x\y)\((x\y)\x) by BCIALG_1:8
      .= x\y by A8,BCIALG_1:2;
    (x\(x\(x\y)))\((x\y)\x) = (x,(x\(x\y))) to_power 1 \ ((x\y)\x) by
BCIALG_2:2
      .= (((x\y),((x\y)\x)) to_power 1,(x\(x\y))) to_power 1 by A7,BCIALG_2:2
      .= (((x\y)\((x\y)\x)),(x\(x\y))) to_power 1 by BCIALG_2:2
      .= ((x\y)\((x\y)\x))\(x\(x\y)) by BCIALG_2:2;
    hence thesis by A10,A9;
  end;
  hence thesis by A6,BCIALG_3:28;
end;
