theorem Th16:
  a = i mod p implies -a = (p-i) mod p
  proof
    assume A1: a = i mod p;
    reconsider b = (p-i) mod p as Element of GF(p) by Th14;
    a+b = (i+(p-i)) mod p by A1,Th15
    .= 0 by INT_1:50
    .= 0.GF(p) by Th11;
    hence thesis by VECTSP_1:16;
  end;
