We have seen that if the number of ancestors of whatever configuration is bounded, all states have the same number of ancestors. Otherwise the number of ancestors would increase as the length of the configuration increased, in agreement with the rate implied by the largest eigenvalue of N. The possibility is still open that there is no upper limit but that the lower limit might not be so small as zero, implying that every configuration has at least one ancestor. This would create a peculiar but not impossible statistical situation wherein an increasing variance did not imply spreading of the data on both sides of the mean.
For automata, a contradiction can be deduced by confronting a configuration with many ancestors with one with minimal ancestors; many ancestors should lead to still more in a composite configuration, but they cannot because of minimality.
Thus, consider any matrix P, some matrix , and a certain abuse of notation with respect to the evolutionary function . Then if m>0 there are words and plus a matrix M, for which
The idea of selecting a fixed Q, varying P, and exploiting the fact that will not work because a counterimage of the bridge M will not necessarily mesh with every . However if the left hand element of P is fixed, an accounting can be made. Therefore let
whereupon it may be concluded that
on account of the fact that these are only some of the items making up the counterimages of the extension QMP. This inequality contemplates all the choices of for a fixed , but there could be as many as choices of . All together, independently of P,
which is the upper bound necessary to conclude that whenever m>0 there are equal numbers of ancestors for all configurations --- exactly . This is the second part of Dubois-Violette and Rouet's  result.
Harold V. McIntosh