reserve x for set;
reserve i,j for Integer;
reserve n,n1,n2,n3 for Nat;
reserve K,K1,K2,K3 for Field;
reserve SK1,SK2 for Subfield of K;
reserve ek,ek1,ek2 for Element of K;
reserve p for Prime;
reserve a,b,c for Element of GF(p);
reserve F for FinSequence of GF(p);

theorem Th27:
  for p be Prime, x be Element of GF(p) st 2 < p & x+x = 0.(GF p)
  holds x = 0.(GF(p))
  proof
    let p be Prime, x be Element of GF(p);
    assume A1:2 < p & x+x = 0.(GF(p));
    x in Segm p; then
    reconsider Ix=x as Element of NAT;
A2: 1= 1.(GF(p)) by Th12;
A3: 1+1 < p by A1;
A4: 1.(GF(p))+1.(GF(p)) = 2 by A2,A3,INT_3:7;
    set d = 1.(GF(p))+1.(GF(p));
A5: d*x = 2*Ix mod p by A4,INT_3:def 10;
    x+x =1.(GF(p))*x +x
    .=1.(GF(p))*x +1.(GF(p))*x
    .= 2*Ix mod p by A5,VECTSP_1:def 7; then
    2*Ix mod p = 0 by A1,Th11; then
    2*Ix - (2*Ix div p) * p= 0 by INT_1:def 10; then
A6: p divides 2*Ix by INT_1:def 3;
    p divides Ix by A1,A6,EULER_1:13,INT_2:28,30; then
    Ix = p * (Ix div p) by NAT_D:3; then
    Ix - (Ix div p)*p = 0; then
A7: Ix mod p = 0 by INT_1:def 10;
    Ix < p by NAT_1:44;
    then Ix = 0 by A7,NAT_D:63;
    hence x = 0.(GF(p));
  end;
