Assume statements S₁ and S₂ defined as: S₁: L₂ − L₁ is recursively enumerable,…

Two closure results govern this question: recursive languages are closed under complement, and a recursively enumerable (r.e.) language intersected with a recursive language is again r.e.; separately, the set of all Turing machines is countable because each machine has a finite description over a finite alphabet.

• Recursive (decidable) languages are closed under complement: if L₁ is recursive, its complement L₁′ is also recursive.

• An r.e. language intersected with a recursive language is r.e.: to recognize membership in the intersection, first run the recursive language's decider, which always halts; only if it accepts do you run the r.e. language's semi-decision procedure — the overall procedure need not halt when the input is not in the intersection, which is exactly what recursive enumerability requires.

• Every Turing machine has a finite transition-table description over a finite alphabet, so it can be written as a finite binary string; the set of all finite binary strings is countably infinite.

Applying this to the two statements:

1. S₁: L₁ is recursive, so its complement L₁′ is recursive. L₂ − L₁ = L₂ ∩ L₁′. Since L₂ is r.e. and L₁′ is recursive, this intersection is r.e. — S₁ holds.

2. S₂: Each Turing machine's description (states, transitions, tape alphabet) is a finite string over a finite alphabet. Enumerating all finite strings by length gives a bijection with a subset of ℕ, so the set of all Turing machines is countable — S₂ holds.

3. Both statements are true, so the correct choice is 'Both S₁ and S₂ are correct.'

The closure-under-complement and r.e.-intersect-recursive results are standard theorems, independent of this specific L₁, L₂ — so S₁ holds for any such pair, not just this one. For S₂, contrast with the (uncountable) set of all possible languages over an alphabet: that larger set is uncountable, which is exactly why not every language is decidable or recognizable — but the machines themselves, being finite descriptions, remain countable.