theorem Th12:
  n = no & m in uparrow no implies n <= m
  proof
    assume that
A1: n = no and
A2: m in uparrow no;
    m in {x where x is Element of NAT:ex p0 be Element of NAT st no = p0 &
            p0 <= x} by A2,CARDFIL2:50;
    then ex x be Element of NAT st m = x & ex p0 be Element of NAT st
           no = p0 & p0 <= x;
    hence thesis by A1;
  end;
