Two Turing machines $M_1$ and $M_2$ with $L(M_1) \subseteq L(M_2)$

Question

Suppose $M_1$ and $M_2$ are two Turing machines such that $L(M_1)\subseteq L(M_2)$. Which of the following is true?

• (A) On every input on which $M_1$ does not halt, M2 does not halt
• (B) On every input on which $M_1$ halts, $M_2$ halts too
• (C) On every input which $M_1$ accepts, $M_2$ halts.
• (D)On every input which $M_2$ accepts, $M_1$ halts.

I am confused between B and C. This was an online practice test question.

For (B) my claim is that when $M_1$ is able to decide the language (i.e., $M_1$ halts on every input), then, since it is given $L(M_1) \subseteq L(M_2)$, $M_2$ should also halt and be able to decide on every input.

However, option (C) also looks convincing. If for every input $M_1$ says "yes" in the language, then $M_2$ should also be able to decide for that input.

Please let me know how to approach this correctly.

EDIT: After reading the discussion, am I correctly interpreting the problem If I imagine the figure below

SO, (B) must be the wrong choice according to this right?

Answer 1

"Halt" is not synonymous with "accept", although the latter implies the former. A TM can halt and reject. On the other hand, $L(M)$ stands for the set of words accepted by $M$, not the words it halts for.

Hence, alternative (C) is correct. (B) is incorrect because $M_1$ could halt on and reject a word which is accepted by $M_2$ (i.e., a word in $L(M_2) \setminus L(M_1)$), or for which $M_2$ does not halt, even.

Note the question makes little sense, if any, if we assume $M_1$ and $M_2$ to be deciders since, then, all alternatives are correct, (A) because the premise is always false (since $M_1$ always halts) and the others because the respective implications are always true.