
theorem X1:
for F being Field
for S being Subset of F st S * S c= S & SQ F c= S
holds S /\ (-S) = {0.F} iff not -1.F in S
proof
let R be Field, S be Subset of R;
assume AS: S * S c= S & SQ R c= S;
A0: 1.R in SQ R;
X: now assume A: S /\ (-S) = {0.R};
   now assume -1.R in S;
     then -(-1.R) in -S;
     then 1.R in S /\ -S by XBOOLE_0:def 4,A0,AS;
     hence contradiction by A,TARSKI:def 1;
     end;
   hence not -1.R in S;
   end;
now assume A: not -1.R in S;
  now assume A2: S /\ (-S) <> {0.R};
    B0: 0.R in SQ R;
    then -0.R in -S by AS;
    then 0.R in S /\ (-S) by AS,B0,XBOOLE_0:def 4;
    then A3: {0.R} c= S /\ (-S) by TARSKI:def 1;
    now assume A4: not(ex a being Element of R st
                                a <> 0.R & a in S /\ (-S));
      now let o be object;
        assume o in S /\ (-S);
        then o = 0.R by A4;
        hence o in {0.R} by TARSKI:def 1;
        end;
      hence contradiction by A2,A3,TARSKI:2;
      end;
    then consider a being Element of R such that
    A5: a <> 0.R & a in S /\ (-S);
    A9: a in S & a in (-S) by A5,XBOOLE_0:def 4;
    then A6: -a in -(-S);
    A7: now assume -a = 0.R;
        then --a = 0.R;
        hence contradiction by A5;
        end;
      ((-a)")^2 is square;
      then ((-a)")^2 in SQ R;
      then A8: ((-a)")^2 * (-a) in S * S by AS,A6;
    (-a)" in S
      proof
      ((-a)")^2 * (-a) = (-a)" * ((-a)" *(-a)) by GROUP_1:def 3
                      .= (-a)" * 1.R by A7,VECTSP_1:def 10
                      .= (-a)";
      hence thesis by A8,AS;
      end;
    then A8: a * (-a)" in S * S by A9;
    (-(a")) * (-a) = a * a" by VECTSP_1:10 .= 1.R by A5,VECTSP_1:def 10; then
    a * (-a)" = a * (-(a")) by A7,VECTSP_1:def 10
             .= -(a * a") by VECTSP_1:8
             .= -1.R by A5,VECTSP_1:def 10;
    hence contradiction by A8,A,AS;
    end;
  hence S /\ (-S) = {0.R};
  end;
hence thesis by X;
end;
