theorem Th20:
  Ant(f) = Ant(g) & Ant(f) |= Suc(f) & Ant(g) |= Suc(g) implies
  Ant(f) |= (Suc(f)) '&' (Suc(g))
proof
  assume
A1: Ant(f) = Ant(g) & Ant(f) |= Suc(f) & Ant(g) |= Suc(g);
  let A,J,v;
  assume J,v |= Ant(f);
  then J,v |= Suc(f) & J,v |= Suc(g) by A1;
  hence thesis by VALUAT_1:18;
end;
