Congruences: Make %
sound by restricting cases where we return a constant
#1161
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For
(c1, m1) % k
withk
constant only return a constant whenk
dividesm1
(already here before), i.e. result will be+/- c1%k
and one of the following hold:
c1%k
is0
(sign does not matter)(c1, m1)
is a constant (sign will be correct)This addresses the soundness part of #1156, leaving it open how to become more precise again (by making use of the sign of the left argument where it is known).