theorem Th19:
  ri < li & [li,ri] is Gap of Gi & ri9 < li9 & [li9,ri9] is Gap of Gi implies
  li = li9 & ri = ri9
proof
  assume that
A1: ri < li and
A2: [li,ri] is Gap of Gi and
A3: ri9 < li9 and
A4: [li9,ri9] is Gap of Gi;
A5: li in Gi by A2,Th13;
A6: ri in Gi by A2,Th13;
A7: li9 in Gi by A4,Th13;
A8: ri9 in Gi by A4,Th13;
  hereby
    assume li <> li9;
    then li < li9 or li9 < li by XXREAL_0:1;
    hence contradiction by A1,A2,A3,A4,A5,A7,Th13;
  end;
  hereby
    assume ri <> ri9;
    then ri < ri9 or ri9 < ri by XXREAL_0:1;
    hence contradiction by A1,A2,A3,A4,A6,A8,Th13;
  end;
end;
