Seadet-i Ebediyye - Endless Bliss Fifth Fascicle

e 6x1/2=3, and the share of the one whose fard is sudus would be
6x1/6=1, is based on [3+1=4] four on account of its being a problem
of return. When there is a daughter and a son’s daughter, three
shares belong to the daughter and one share belongs to the son’s
daughter.
A third example: If the problem of return contains the
categories of nisf and thuluth or sudusân [two units of sudus] and
nisf or thuluthân [two units of thuluth] and sudus, the problem is
based on five instead of six. When there is one sister and two
sisters uterine the basis of the problem of return becomes [3+2=5]
five, three shares being given to the sister and one to the two sisters
uterine.
2 - When the problem of return contains the unreturned too,
there are, again, two possible situations:
A: When the returned have their fards in (the same) one
category, either one of two cases exist(s):
First case: If, after the unreturned has gotten his (or her) share,
the remaining property can be divided by the number of the
returned, the unreturned gets his (or her) share and the rest is
divided by the number of the returned.
For example, when there is a husband and three daughters, the
husband gets one of the four shares and the remaining three shares
are distributed to the daughters.
Second case: If, after the unreturned has gotten his (or her)
share, the remaining property cannot be divided by the number of
the returned, the basis of the problem is found by multiplying the
number of the persons returned by the denominator of the fard
[share] of the unreturned.
For example, when there is a husband and five daughters, the
husband gets the rubu’ (one-fourth) and the remaining three
shares cannot be divided between the five daughters; so the basis
of the problem becomes [4x5=20] twenty, the husband gets five
shares and the daughters get fifteen shares; each daughter is given
three shares.
B: If the returned own fards in two or three different
categories, the person unreturned gets his (or her) share and the
remaining property is divided like in the problem of return. Here
also, there are two cases:
First case: If the shares remaining from the unreturned can be
divided by the basis of the problem of return, for determining the
basis of the problem the least common multiple of the number of
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