Sound integral enclosure from certified adaptive Simpson

Fix a recursive panel $[u,v]$ of length $L$ with finite oracle results. By [A1], the exact values $f(u),f((u+v)\mathrm{/}2),f(v)$ belong to their returned intervals. The interval $W=[v,v]-[u,u]$ in [F4] contains $L=v-u$, and $[6,6]$ is an exact positive divisor. Thus the expression DAG for $S_P$ in the algorithm contains the exact $L(f)$ expression of [F1]; by [F3]–[F4], $S_P$ contains the exact Simpson value $S$.

By [A1], every value of $f^{(4)}$ on $[u,v]$ lies between $y_{-}$ and $y_{+}$; ordered endpoints [F4] imply its absolute value is at most $M=\max (\mathrm{\mid }{y}_{-}\mathrm{\mid },\mathrm{\mid }{y}_{+}\mathrm{\mid })$. This $M$ is nonnegative. By [F6], the restriction of $f$ to $[u,v]$ is $C^4$ and $g:=f^{(4)}$ is continuous. For any $x,y\in [u,v]$, [F11] gives $\mid \mathrm{\mid }g(x)\mathrm{\mid }-\mathrm{\mid }g(y)\mathrm{\mid }\mid \le \mathrm{\mid }g(x)-g(y)\mathrm{\mid }$. Fix $x_0\in [u,v]$ and let $\epsilon >0$. Continuity of $g$ at $x_0$ supplies $\delta >0$ such that $x\in [u,v]$ and $\mathrm{\mid }x-x_0\mathrm{\mid }<\delta$ imply $\mathrm{\mid }g(x)-g(x_0)\mathrm{\mid }<\epsilon$. Then [F11] gives $\mid \mathrm{\mid }g(x)\mathrm{\mid }-\mathrm{\mid }g(x_0)\mathrm{\mid }\mid <\epsilon$, so $\mathrm{\mid }g\mathrm{\mid }$ is continuous at $x_0$. Since $x_0$ was arbitrary, $\mathrm{\mid }g\mathrm{\mid }$ is continuous on $[u,v]$. By [F9], $[u,v]$ is compact, and [F10] gives $M_4={\max }_{x\in [u,v]}\mathrm{\mid }g(x)\mathrm{\mid }$. Since every $\mathrm{\mid }g(x)\mathrm{\mid }\le M$ by [A1] and [F4], $M_4\le M$. Applying [F2] with $n=2$ and step $h=L\mathrm{/}2$ gives $\mathrm{\mid }{\int }_u^{v}f-S\mathrm{\mid }\le M_4{L}^5\mathrm{/}2880\le M{L}^5\mathrm{/}2880$. The directed upward rounding in the algorithm makes $B_P\ge M{L}^5\mathrm{/}2880$ by [F5]. Thus the exact panel integral belongs to $S+[-B_P,B_P]$. Outward interval addition and [F3] show that it belongs to $C_P$.

For $p=1$ or $p=2$, [F8] forbids a split, so after the finite panel calculation the call returns. For $p\ge 3$, a call either returns without splitting or splits into two adjacent halves with positive budgets $p_L,p_R<p$ and $p_L+p_R=p-1$, by [F8]. Strong induction on $p$ therefore proves termination; a split tree uses at most $1+p_L+p_R=p$ calls, so the public call uses at most $P$.

The returned panels form a finite partition of $[a,b]$. Each leaf interval contains its exact panel integral by step 1.2; [F7] adds those integrals to $I$. At each internal node the outward interval sum contains the exact sum of its child interval values by [F3] and [F8]. Induction up the finite tree proves $I\in C$.

Since $C=[L,U]$ contains $I$, its midpoint and half-width satisfy $\mathrm{\mid }I-(L+U)\mathrm{/}2\mathrm{\mid }\le (U-L)\mathrm{/}2$. By [F8], the algorithm returns certified only after the exact comparison $r\le \tau$; otherwise it returns inconclusive. An explicit oracle failure, malformed or non-finite oracle interval, or non-finite interval operation returns no finite enclosure by definition. The truth of the oracle's enclosure contract remains a hypothesis, not a runtime test.