$1.\: A_{DFA} = \{\langle B, w \rangle \mid B \text{ is a } DFA \text{ that accepts input string } w \}$ $2.\:A_{DFA} = \{\langle B \rangle \mid B \text{ is a } DFA \text{ that accepts input string } w \}$

I know how to proof that 1 is decidable by constructing a machine that always halts and accepts whenever $B$ accepts, otherwise rejects. what's the difference between 1 and 2. I find that it doesn't make sense to not include $w$ inside the encoding brackets but I have seen this notation in other places. are they both the same?


1 Answer 1


The languages are not the same.

In the first one, $w$ is a part of the input.

In the second one, $w$ is fixed beforehand, and the language has to depend on what you fix it to be.

  • $\begingroup$ Will that affect the proof that $A_{DFA}$ is decidable. by other means, How is it possible to proof that a language is decidable if $w$ is not part of the input? $\endgroup$
    – abeer00
    Commented Apr 11, 2021 at 19:08
  • $\begingroup$ It definitely can only help if $w$ is fixed beforehand and is not a part of the input, but in any case both languages are decidable since we can emulate the DFA until it halts (and it always halts) $\endgroup$
    – nir shahar
    Commented Apr 11, 2021 at 22:12

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