There is no recursive or definable $R$ such that ......
The answer depends on how broadly or narrowly the term `matrix method' is defined.
If ...... then $R$ is right Noetherian provided $R$ is semiprime [2] or commutative [4] or $R/N$ has zero socle.
The case when $f$ is decreasing can be proved similarly, or else can be deduced from ......
Then $f=g$, or equivalently $a(f)=a(g)$.
Here the interesting questions are not about individual examples, but about the asymptotic behaviour of the set of examples as one or another of the invariants (such as the genus) goes to infinity.
Moreover, for $L$ tame or otherwise, it may happen that $E$ is a free module.
Then $F$ may or may not fix $B$.
The question we shall be concerned with is whether or not $f$ is ......
Its role is to rule out having two or more consecutive $P$-moves.
Any vector with three or fewer 1s in the last twelve places has at least eight 1s in all.
The intervals we are concerned with are either completely inside $A$ or completely inside $B$.
By Corollary 2, distinct 8-sets have either zero, two or four elements in common.
Either $f$ or $g$ must be bounded.
Any map either has a fixed point, or sends some point to its antipode.
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