theorem Th65:
  r <= a & a < [\r/]+1 implies [\a/] = [\r/]
proof
  assume
A1: r <= a;
  assume a < [\r/]+1;
  then
A2: a-1 < [\r/]+1-1 by XREAL_1:9;
  [\r/] <= r by Def6;
  then [\r/] <= a by A1,XXREAL_0:2;
  hence thesis by A2,Def6;
end;
