Conditional Concatenated Threshold for Proper Codes

Since $q=\lfloor d\mathrm{/}(2\mathrm{\ell })\rfloor$, we have $2q\mathrm{\ell }\le d$. Also $A\ge 1$, and $A{\eta }^q<1$ gives $\theta :=A^{1\mathrm{/}q}\eta <1$. Apply [F2] to obtain $p_r\le A^{-1\mathrm{/}q}{\theta }^{(q+1{)}^r}$.

If $\eta =0$, the product fault law gives probability zero to every nonempty fault pattern, so $p_0=0$ and level zero suffices. If $\eta >0$, put $c=-\log \theta >0$. Any $r$ with $(q+1{)}^r\ge c^{-1}\log (N{A}^{-1\mathrm{/}q}\mathrm{/}\delta )$ gives $N{p}_r\le \delta$ by [F2]; if the logarithm is nonpositive or its positive value is at most $c$, level $r=0$ suffices. For fixed $A,q,\eta$ with $0<\theta <1$, the least qualifying level is $O(\log \log (N\mathrm{/}\delta ))$ as $N\mathrm{/}\delta \to \mathrm{\infty }$.

The event that any of the $N$ top-level circuit rectangles is bad is the union of $N$ events, each of probability at most $p_r$. The finite union bound [F8] gives $\alpha :=\mathbb{P}(\text{some top-level rectangle is bad})\le N{p}_r$.

For every fault pattern, [F9] gives a conditional channel, so its output on the fixed joint data input and external reference is a density operator. On the event that all top-level rectangles are $q$-good, recursive goodness gives at most $q$ bad children in each computation and correction stage. The one-level contracts [F7], finite hypothesis [A1], and simultaneous recursive induction [F3] place the final corrected instrument in its hierarchical residual sector. Applying its recursive decoder $D_r$ returns the ideal logical instrument exactly, so every such pattern has decoded output ${\rho }_{\mathrm{i}\mathrm{d}}$. If $\alpha >0$, the normalized finite mixture of outputs from bad patterns is a density operator $\tau$; if $\alpha =0$, choose any density operator for $\tau$. Thus ${\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}}=(1-\alpha ){\rho }_{\mathrm{i}\mathrm{d}}+\alpha \tau$, and [F10] confirms that this weighted schedule output is a channel.

By homogeneity of the trace norm, $D({\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}},{\rho }_{\mathrm{i}\mathrm{d}})=\alpha D(\tau ,{\rho }_{\mathrm{i}\mathrm{d}})\le \alpha \le N{p}_r$, where the first inequality is [F5]. The same identity holds after tensoring any reference identity, since the gadget contracts preserve the external reference after internal outcomes and fault memory are traced.

Apply [F6] to the final finite POVM. Its output total-variation distance is at most $D({\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}},{\rho }_{\mathrm{i}\mathrm{d}})\le N{p}_r$. For any success event $E$, $\mathrm{\mid }{P}_{\mathrm{o}\mathrm{u}\mathrm{t}}(E)-P_{\mathrm{i}\mathrm{d}}(E)\mathrm{\mid }\le \frac{1}{2}{\sum }_x\mathrm{\mid }{P}_{\mathrm{o}\mathrm{u}\mathrm{t}}(x)-P_{\mathrm{i}\mathrm{d}}(x)\mathrm{\mid }$, so the success-probability change is at most $N{p}_r$.

Because the gadget family is fixed and finite, choose a constant $\kappa \ge 1$ bounding both the number of elementary locations and the scheduled depth of every replacement gadget, including preparation, control, reset, and idle locations. Replacing each location of a schedule of size $S$ and depth $D$ gives size at most $S\kappa$ and depth at most $D\kappa$: replace disjoint locations of each original layer in parallel and pad their gadgets to the common depth bound. Induction gives size at most $S{\kappa }^r$ and depth at most $D{\kappa }^r$ at level $r$. Thus the multiplicative size and depth overheads are at most ${\kappa }^r$, which is a fixed power of $\log (N\mathrm{/}\delta )$ by step 1.2.