theorem Th4:
  b1 <= b2 iff seq(a,b1) c= seq(a,b2)
proof
  thus b1 <= b2 implies seq(a,b1) c= seq(a,b2)
  proof
    assume b1 <= b2;
    then
A1: b1+a <= b2+a by XREAL_1:6;
    let b be object such that
A2: b in seq(a,b1);
    reconsider c = b as Element of NAT by A2;
    c <= b1+a by A2,Th1;
    then
A3: c <= b2+a by A1,XXREAL_0:2;
    1+a <= c by A2,Th1;
    hence thesis by A3;
  end;
  assume
A4: seq(a,b1) c= seq(a,b2);
  now
    assume b1 <> 0;
    then b1+a in seq(a,b1) by Th3;
    then b1+a <= b2+a by A4,Th1;
    hence thesis by XREAL_1:6;
  end;
  hence thesis;
end;
