theorem
  i > 0 & i+m in Seg (n+m) implies i in Seg n & i in Seg (n+m)
  proof
  assume that
A1: i > 0 and
A2: i+m in Seg (n+m);
  1 = 0+1;
  then
A3: 1 <= i by A1,NAT_1:13;
A4: i+m <= n+m by A2,Th1;
  i <= n by A2,Th1,XREAL_1:6;
  hence i in Seg n by A3;
  i <= i+m by NAT_1:11;
  then i <= n+m by A4,XXREAL_0:2;
  hence thesis by A3;
end;
