Let me by means of an example introduce the reader into how a deduction of a stem is carried out in practice. I’ll choose a signgroup at random "B13", and then I intend to investigate its contents of stems, if any.
A stem is a construction of two and only two signs. By this means, there are given three compositions for a signgroup with five signs as B13: 1) ØS RR F. 2) ØS R RF. 3) Ø SR RF. Taking the potential stems in the first composition, beginning
with RR, this combination of signs is to be searched for in all 61 signgroups. - It is not found. The search proceeds with ØS - the result is still negative. Only the combination SR in the third composition has a repetition; namely in the signgroup
B05. B05 consist of four signs, and by that it has two compositions: B SR K and BS RK. The stem SR is yet not guaranteed before both compositions of B05 have been tested. The first SR was tried with a positive result. The potential stems in the second composition
cannot be recovered. The stem SR is by now determined. On the other hand, if the possible second compositions BS or RK were recovered, but only at one location, just as SR, then the contents of stems in B13 and B05 would have been evasive, cf. B22, B29 and
A04. Thus, a majority in favour of the deduced stem is demanded before certainty is attained. In A17, A29 and A23, the combination WB emerges, but it is overlapping the stem BA, which is present thirteen times in all, by means of which WB is not to be a stem.
The two identical signgroups A17 and A29 must, besides BA, of necessity hold two more pairs of stems, but a search determines that none of the compositions available are found in other signgroups, without which the stems cannot be deduced with certainty. At
this point the outcome must be based on an estimate: That the position for a sign in a stem is to be respected, when it is known from a definable stem, due to the definable stems observing these position rules among themselves.