Recall from Theorem 3 that there is a sequence $(a_n)$ of elements of $U$ that is cofinal in $M$.
Altering finitely many terms of the sequence $(u_n)$ does not affect the validity of (9).
Let $(a_n)$ be the sequence of zeros of $f$ arranged so that $|a_1|\leq|a_2|\leq$ ......
Extend this sequence of numbers backwards, defining $N_{-1}$, $N_{-2}$ and $N_{-3}$ by ......
Then the sequence (8) breaks off in split exact sequences.
Thus the long exact sequence breaks up into short sequences.
The exact sequence ends on the right with $H(X)$.
The proof proper [= The actual proof] will consist of establishing the following statements in sequence.
the all-one sequence
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