The safeguarded Euler pressure solver preserves its root bracket and terminates

At $k=0$, [F2] initializes $[0,P]$ with $\mathrm{\Phi }(0+)<0<\mathrm{\Phi }(P)$ by [F1], so it contains $p_{\ast }$. This establishes the initial bracket.

Assume the current bracket contains $p_{\ast }$ with nonpositive left and nonnegative right residuals. A midpoint query lies in its central half; any accepted Newton query is in the central quarter, and all other Newton outcomes use the midpoint. By strict increase in [F1], a negative query residual leaves the root to its right and a positive one to its left, so the corresponding endpoint replacement preserves the bracket; an exact zero returns the singleton. Every nonexact query is at least one quarter of the old width from either endpoint, so the new width is at most $3\mathrm{/}4$ of the old width.

Induction on completed nonexact queries using the base step 1.1 and the preserved-bracket transition in step 1.2 gives containment and width at most $(3\mathrm{/}4{)}^{k}P$. Each outer pass calls the Newton procedure with budget $1$, so [F3] supplies at most one correction and finitely many scalar residual evaluations; the one-dimensional correction is finite by [F4], and all primitive operations terminate by [A1]. At most $N$ passes therefore return one of the defined statuses.

If $(3\mathrm{/}4{)}^{N}P\le \tau$, the width test in [F2] succeeds no later than the final check, so the algorithm returns a pressure bracket unless it already returned an exact singleton. Otherwise a budget-exhausted result retains the bracket invariant from step 2.1.