P/poly equals polynomial time with polynomial advice

Suppose $(C_n)$ has size at most $p(n)$. Encode $C_n$ as a topologically ordered gate list: each of its at most $p(n)$ records contains a constant-size gate tag and at most two indices of $O(\log (n+p(n)))$ bits. This is polynomial advice. A fixed evaluator scans the list in topological order and computes $C_n(x)$ in polynomial time, so [L2] gives an advised decider.

Conversely, let $M(x,a_n)$ run in polynomial time with $\mathrm{\mid }{a}_n\mathrm{\mid }\le p(n)$. Use a fixed self-delimiting encoding $⟨x,a⟩$ whose length depends only on $\mathrm{\mid }x\mathrm{\mid }$ and $\mathrm{\mid }a\mathrm{\mid }$, and let $B=\{⟨x,a⟩:M(x,a)=1\}$. This is a language in $\mathrm{P}$, so [L3] supplies polynomial-size circuits for $B$. For a fixed $n$, put $m_n=\mathrm{\mid }{a}_n\mathrm{\mid }$ and take the circuit for the common encoding length of $⟨x,a_n⟩$ with $\mathrm{\mid }x\mathrm{\mid }=n$. Hardwire the delimiter bits and all bits of $a_n$, leaving only the $n$ bits of $x$ free. Because $m_n\le p(n)$, the resulting $n$-input circuit has polynomial size and agrees with the original computation $M(x,a_n)$; no change to the advice string or its length is made.

Step 1.1 proves the circuit-to-advice inclusion and step 1.2 proves the advice-to-circuit inclusion, including the advice-length and simulation-size bounds.