From http://rjlipton.wordpress.com/2009/05/27/arithmetic-hierarchy-and-pnp/,
Define, $M_{[x,c]}$ as the deterministic Turing machine that operates as follows on an input $y$. The machine treats $x$ as a deterministic program, and simulates $x$ on input $y$. At the same time the machine runs a counter that stops its execution after steps $|y|^c$. If the machine accepts before the counter stops, then it accepts; otherwise, it rejects.
Let $f(i,c)$ be the smallest natural number so that $M_{[i,c]}$makes a mistake on the input $y$. Then, if $P \neq NP$ is true, the function $f(i,c)$ is always defined.
Theorem: Suppose that there are infinite number of $i$ for which there exists a $c$ so that $$f(i,c) > 2^{2^{|i|+c}}$$ Then, for infinitely many $n$, SAT has circuit size $n^{O(\log n)}$.
Proof: Let $i>1$ and $c$ be so that $$f(i,c) > 2^{2^{|i|+c}}$$ Define $n = 2^{|i|+c-1}$. Note, that $c$ is at most $\log n$. Then, $M_{[i,c]}$ on all $y$ of length $n$ is correct, since $y \leq 2^n = 2^{2^{|i|+c-1}} < f(i,c)$. The size of the circuit that simulates this Turing machine on inputs of length $n$ is polynomial in $|i|$, $n$, and the running time of the machine. The machine, by definition, runs in time $|y|^c \leq n^c \leq n^{\log n}$
I am not getting this part. Can anyone explain this (to specify, “The size of the circuit that simulates this Turing machine on inputs of length $n$ is polynomial in $|i|$, $n$, and the running time of the machine” in the quote)? (So the question is how can we relate the running time of Turing machine to the size of the circuit.)