Assume $L_1$, $L_2$, and $L_3$ are recursively enumerable languages. Prove that $L = (L_1L_2) \cap L_3$ is a recursively enumerable language by designing a non-deterministic Turing machine that semi-decides $L$ utilizing Turing machines for $L_1$, $L_2$, and $L_3$.
Added by Jesse D.
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First, let's define the language L1L2 as the concatenation of strings in L1 and L2. This means that for any string w in L1L2, there exist strings x in L1 and y in L2 such that w = xy. Now, let's design the non-deterministic Turing machine M for L: Show more…
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(a) For every language L over Σ at least one of L and L^C is Turing-recognizable. (b) If languages L_1 and L_2 over Σ are both regular with L_1 ⊆ L_2, then a language L that is sandwiched between them, that is L_1 ⊆ L ⊆ L_2 is Turing-recognizable. (c) For every language L over Σ that is Turing-recognizable, at least one of L and L^C is Turing-decidable. Prove the following statement: (d) Let T_1 and T_2 be Turing Machines such that L(T_2)^C ⊆ L(T_1). If, on every input word w, T_1 takes longer to compute than T_2, then L(T_2) is Turing-decidable.
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