Answer :
The statement that for all languages L1, L2 over σ, if L1 ⊆ L ⊆ L2 and L1, L2 are decidable, then L is also decidable is proven true.
To prove that L is decidable given that L1 ⊆ L ⊆ L2 and L1, L2 are decidable, we need to understand what it means for a language to be decidable and how deciders for L1 and L2 can be utilized to show that L is also decidable.
Decidable Language:
A language is said to be decidable if there exists a Turing machine (which we can think of as an algorithm) that can accept any input string and, after a finite amount of time, halt and correctly decide whether the string belongs to the language or not.
Given that:
1. L1 and L2 are decidable.
2. L1 ⊆ L ⊆ L2.
We know that for L1 and L2, there exist two Turing machines: one for L1, which we can call T1, and one for L2, which we can call T2. T1 and T2 will each take a string as input and decide whether that string is a member of L1 or L2, respectively.
Given a string 's', to decide whether 's' belongs to L, we can perform the following steps:
Step 1:
Run s on T1 (the decider for L1).
- If T1 rejects 's', then 's' is not in L1. Given that L1 ⊆ L, if 's' is not in L1, then it also cannot be in L. Hence, 's' is not in L, and we can reject 's'.
- If T1 accepts 's', then 's' is in L1, and since L1 ⊆ L, 's' could potentially be in L. We must check further, so we proceed to the next step.
Step 2:
Run s on T2 (the decider for L2).
- If T2 accepts 's', then 's' is in L2. Given that L ⊆ L2, if 's' is in L2, it could also be in L. Since we know from step 1 that 's' is in L1, and L1 ⊆ L, we can now decide that 's' is indeed in L. Therefore, we can accept 's'.
- If T2 rejects 's', then 's' is not in L2. Since all strings in L are also in L2 (because L ⊆ L2), if 's' is not in L2, it cannot be in L. Hence, 's' is not in L, and we can reject 's'. (However, this scenario will not occur in this setup because we have already established that 's' is in L1 and L1 ⊆ L2, implying 's' will always be in L2.)
Since the result of T1 already ensures that 's' would be accepted by T2 (because L1 ⊆ L2), we only need to run T1 to decide on rejection, and in case of T1 accepting 's', we validate the membership in L using T2.
Because we are able to reach a decision for any string 's' using these steps, combining the deciders for L1 and L2, we have created a decider for L. Consequently, we have shown that L is decidable.
Hence, the statement that for all languages L1, L2 over σ, if L1 ⊆ L ⊆ L2 and L1, L2 are decidable, then L is also decidable is proven true.
