The $n$th slice of a set $A \subseteq \Sigma^{*}$ is defined as:

$$A_n = \{x \in \Sigma^{*}\mid\langle n,x\rangle \in A\}$$

The definition of parameterization is as follows - $C$ parameterizes $D$ (also called $C$ is universal for $D$) if,

$$\exists A \in C \quad\text{s.t}\quad D = \{A_n|n\in \mathbb{N}\}$$

However, I have no idea how to prove that the class of decidable languages (DEC) can parameterize DEC.

  • $\begingroup$ What are $A_n$ here? And what is $A$ used for? $\endgroup$
    – nir shahar
    Mar 7, 2022 at 20:48
  • $\begingroup$ @nirshahar A is a finite sequence of strings. I have updated the question with the definition of An above. $\endgroup$ Mar 7, 2022 at 20:53
  • $\begingroup$ Thanks. How is $\langle n,x \rangle$ defined here? Are we working with binary alphabet (i.e, $\Sigma=\{0,1\}$) and $n$ is interpreted as a binary number? $\endgroup$
    – nir shahar
    Mar 7, 2022 at 21:12
  • $\begingroup$ @nirshahar n is a natural number. So to determine if a string is in A_n, we stick the number n in front of it and check it is in A. $\endgroup$ Mar 8, 2022 at 1:52
  • 1
    $\begingroup$ But some natural numbers may not even be a part of your alphabet! This would make a syntactically problematic question here. If $\Sigma=\{a,b\}$, then what is $5a$ here? It is not part of $\Sigma^*$, since $5\notin \Sigma$... $\endgroup$
    – nir shahar
    Mar 8, 2022 at 10:08

1 Answer 1


Okay first you need to know that there exists a numbering $\phi$ for all the Turing-Machines that is $\phi: \mathbb{N} \to TM$ is a computable surjection. This is often called discription number. Now we define: $$ A = \{\langle p,w \rangle | \text{ the TM } \phi(p) \text{ accepts } w\} $$ This language is decidable since $\phi$ is computable and using the Universal TM we can simulate $\phi(p)$ on $w$. Thus $A \in DEC$. But now $$ \{A_n|\, n\in \mathbb{N}\} = DEC $$ because for every TM $B$ there exists a $n\in \mathbb{N}$ such that $\phi(n)$ is an encoding of $B$. Hence $A_n$ is the language decided by $B$.


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