Claim : $\cup_{c,d} $ DTIME$(n^c)/n^d \subseteq$ $P_{poly}$

Proof : if $L$ is decidable by a polynomial-time Turing machine $M$ with access to advice family $\{\alpha_n\}_{n\in \mathbb{N}}$ of size $a(n)$, then we can use the cook-levin construction for every $n$ a polynomial-sized circuit $D_n$ such that on every $x \in \{0,1\}^n$, $\alpha_n \in \{0,1\}^{a(n)}$, $D_n(x,\alpha) = M(x,\alpha)$. That is, $C_n$ is equal to the circuit $D_n$ with the string $\alpha_n$ "hard-wired" as its second input.

Question : I am not getting the part that circuit is going to have two inputs one hard-wired as second input. How a circuit will like with two inputs. I have seen circuits with one input but not with two inputs (i.e. one input is hard-wired). I am thinking that it may be the case that they are attaching advice bit to each of AND and OR gate (increasing the fan-in).

Reference : http://theory.cs.princeton.edu/complexity/book.pdf

Claim : $\cup_{c,d} $ DTIME$(n^c)/n^d \subseteq$ $P_{poly}$

Proof : if $L$ is decidable by a polynomial-time Turing machine $M$ with access to advice family $\{\alpha_n\}_{n\in \mathbb{N}}$ of size $a(n)$, then we can use the cook-levin construction for every $n$ a polynomial-sized circuit $D_n$ such that on every $x \in \{0,1\}^n$, $\alpha_n \in \{0,1\}^{a(n)}$, $D_n(x,\alpha) = M(x,\alpha)$. That is, $C_n$ is equal to the circuit $D_n$ with the string $\alpha_n$ "hard-wired" as its second input.

Question : I am not getting the part that circuit is going to have two inputs one hard-wired as second input. How a circuit will like with two inputs. I have seen circuits with one input but not with two inputs (i.e. one input is hard-wired). I am thinking that it may be the case that they are attaching advice bit to each of AND and OR gate (increasing the fan-in). So I am not getting, how they are hard-wiring the advice string ?

Reference : http://theory.cs.princeton.edu/complexity/book.pdf