reserve            x for object,
               X,Y,Z for set,
         i,j,k,l,m,n for Nat,
                 r,s for Real,
                  no for Element of OrderedNAT,
                   A for Subset of [:NAT,NAT:];

theorem Th11:
  n = no & n <= m implies m in uparrow no
  proof
    assume that
A1: no = n and
A2: n <= m;
    reconsider m as Element of NAT by ORDINAL1:def 12;
    reconsider p0 = no as Element of NAT;
    m in {x where x is Element of NAT:ex p0 be Element of NAT st
                 no = p0 & p0 <= x} by A1,A2;
    hence thesis by CARDFIL2:50;
  end;
