theorem Th19:
  for gi, gj, a be Element of GF(p) st
  gi = i mod p & gj = j mod p & j = i + 1 holds
  gi*a + a = gj*a
  proof
    let gi, gj, a be Element of GF(p) such that
    A1: gi = i mod p & gj = j mod p & j = i + 1;
    reconsider g1 = 1 mod p as Element of GF(p) by EC_PF_1:14;
    p > 1 by INT_2:def 4;
    then g1 = 1 by NAT_D:63
    .= 1.GF(p) by EC_PF_1:12;
    then gi*a + (1.GF(p))*a = gj*a by A1,Th18;
    hence thesis;
  end;
