theorem Th19:
  len f > 1 & Ant(f) = Ant(g) & 'not' p = Suc(Ant(f)) & 'not' Suc(
  f) = Suc(g) & Ant(f) |= Suc(f) & Ant(g) |= Suc(g) implies Ant(Ant(f)) |= p
proof
  assume that
A1: len f > 1 and
A2: Ant(f) = Ant(g) and
A3: 'not' p = Suc(Ant(f)) and
A4: 'not' Suc(f) = Suc(g) & Ant(f) |= Suc(f) & Ant(g) |= Suc(g);
A5: len Ant(f) > 0 by A1,Th4;
A6: now
    given A,J,v such that
A7: J,v |= Ant(f);
    J,v |= Suc(f) & J,v |= 'not' Suc(f) by A2,A4,A7;
    hence contradiction by VALUAT_1:17;
  end;
  let A,J,v such that
A8: J,v |= Ant(Ant(f));
  now
    assume J,v |= Suc(Ant(f));
    then J,v |= Ant(f) by A5,A8,Th17;
    hence contradiction by A6;
  end;
  hence thesis by A3,VALUAT_1:17;
end;
