Is the complement of a Non-Recognizable language

  1. Recognizable
  2. Non-Recognizable
  3. May be Recognizable, May be Non-recognizable. I.e cant comment.

A mathematical proof would be of great help since im unable to think of any way to prove this.

I did some research on this and found below examples. Im specifying complement using a "!"

  1. !A(TM) is a non-recognizable language while A(TM) is a recognizable language.
  2. EQ(TM) is a non-recognizable language and !EQ(TM) is also non-recognizable language

The above two would mean that we simply cant comment on the Recognizability of the complement of non-recognizable language. But I feel that there should be some way to prove (or disprove) this.


1 Answer 1


Assuming that you can prove statement 1 and 2 above, you have just presented a proof. You're trying to show that there exists a language which is non-recognizable such that its complement is recognizable. You then prove that $!A(TM)$ has this property. Then you want to show that there exists a non-recognizable language whose complement is also non-recognizable and then you show that $EQ(TM)$ has this property.

The point is that to prove that you can't comment, you need only to show that both possibilities are possible ie. that examples of each exist.

  • 1
    $\begingroup$ Just a note: point 1 is a direct consequence of the easy lemma: $L$ is decidable if and only if $L$ and $\bar{L}$ are recognizable. Point 2 can be proved using a Turing reduction from $EQ_{TM}$ to $\overline{A_{TM}}$ and from $\overline{EQ_{TM}}$ to $\overline{A_{TM}}$ $\endgroup$
    – Vor
    Oct 19, 2012 at 20:01

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