[see also: have to, necessarily, force]
Any algorithm to find max must do at least $n$ comparisons.
We must have $Lf=0$, for otherwise we can replace $f$ by $f-Lf$.
Our present assumption implies that the last inequality in (8) must actually be an equality.
If there are to be any nontrivial solutions $x$ then any odd prime must satisfy ......
In outline, the argument follows that of the single-valued setting, but there are several significant issues that must be addressed in the $n$-valued case.
Nevertheless, in interpreting this conclusion, caution must be exercised because the number of potential exceptions is huge.
Theorem 3 may be interpreted as saying that $A=B$, but it must then be remembered that ......
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