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 Th50:
  X is BCK-implicative BCK-algebra iff X is BCK-algebra of 1,0,0,0
proof
  thus X is BCK-implicative BCK-algebra implies X is BCK-algebra of 1,0,0,0
  proof
    assume
A1: X is BCK-implicative BCK-algebra;
    for x,y being Element of X holds Polynom (1,0,x,y) = Polynom (0,0,y,x)
    proof
      let x,y be Element of X;
A2:   (x\(x\y))\(x\y) = y\(y\x) by A1,BCIALG_3:35;
      ((x,(x\y)) to_power (1+1),(y\x)) to_power 0 = (x,(x\y)) to_power 2
      by BCIALG_2:1
        .= (x\(x\y))\(x\y) by BCIALG_2:3
        .= (y,(y\x)) to_power 1 by A2,BCIALG_2:2
        .= ((y,(y\x)) to_power 1,(x\y)) to_power 0 by BCIALG_2:1;
      hence thesis;
    end;
    hence thesis by A1,Def3;
  end;
  assume
A3: X is BCK-algebra of 1,0,0,0;
  for x,y being Element of X holds (x\(x\y))\(x\y) = y\(y\x)
  proof
    let x,y be Element of X;
A4: Polynom (1,0,x,y) = Polynom (0,0,y,x) by A3,Def3;
    (x\(x\y))\(x\y) = (x,(x\y)) to_power 2 by BCIALG_2:3
      .= ((x,(x\y)) to_power (1+1),(y\x)) to_power 0 by BCIALG_2:1
      .= (y,(y\x)) to_power 1 by A4,BCIALG_2:1
      .= y\(y\x) by BCIALG_2:2;
    hence thesis;
  end;
  hence thesis by A3,BCIALG_3:35;
end;
