On an old worksheet I came across the question

If L1 and L2 are two Turing decidable languages, then show that 𝐿1∪𝐿2 and 𝐿1𝑜𝐿2 are Turing decidable languages (high-level description with stages is enough).

How do I go about answering this without being given a language to work from?


1 Answer 1


Both $L_1$ and $L_2$ are decidable. Hence, they have algorithms $A_1$ and $A_2$ (respectively) that decide them.

Try to create a new turing machine (algorithm) using the two algorithms $A_1$ and $A_2$.

For example, for the union $L_1\cup L_2$, you can create the following algorithm:

  • run $A_1$ on the input. If it accepted, then also accept.
  • else, run $A_2$ on the input, and return what $A_2$ returned.
  • $\begingroup$ I was able to understand this and I was able to do the same if it was intersection, but how does this work for concatenation? Thanks by the way! $\endgroup$ Commented Apr 26, 2021 at 21:17
  • $\begingroup$ Try to enumerate on all prefixes $\endgroup$
    – nir shahar
    Commented Apr 26, 2021 at 21:35
  • $\begingroup$ For each enumeration check if each of the two parts are in $L_1$ and $L_2$ respectively $\endgroup$
    – nir shahar
    Commented Apr 26, 2021 at 22:03

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