theorem
  i in Seg n implies i+m in Seg (n+m)
proof
  assume
A1: i in Seg n;
  then
A2: 1 <= i by Th1;
A3: i <= n by A1,Th1;
  i <= i+m by NAT_1:11;
  then 1 <= i+m by A2,XXREAL_0:2;
  hence thesis by A3,Th1,XREAL_1:7;
end;
