Finite-budget SCF response iterations return a status in bounded outer work

At the initial entered index $0$, no update has occurred. Each baseline procedure permits at most one residual test after its spectrum and gap checks; the certified DIIS procedure permits one current residual test and, only if $0<N$, at most one candidate residual test.

Before an entered index $k\le N$, its preceding entered indices are exactly $0,\dots ,k-1$, with at most $k$ preceding updates, at most $k$ residual tests for either baseline procedure, and at most $2k$ residual tests for the certified DIIS procedure.

A baseline pass permits at most one residual test; a certified DIIS pass permits at most two, for its current point and a candidate. A continuing pass requires $k<N$ and makes at most one update before entering $k+1$. Thus the preceding counts increase by at most one update and, respectively, one or two residual tests, proving step 1.2 at the next index. Any earlier gap, eigensolver, response-evaluation, invariance, residual or budget return creates no further pass.

The base and induction step show that entered indices form an initial segment of $0,\dots ,N$, with at most $N+1$ residual tests for either baseline procedure and at most $2N+1$ for certified DIIS, since index $N$ checks the current response and returns before a candidate attempt or update. All three make at most $N$ updates. Every entered pass has finitely many returning primitives, so each procedure terminates with a stated status, including when $N=0$. These are only outer-work and status bounds.