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Which of these statements are true for all choices of TM $M$, string $w$, and language $L$?

  1. If $M$ recognizes $L$ and $M$ rejects $w$, then $w\notin L$.
     
  2. If $M$ recognizes $L$ and $w\notin L$, then $M$ rejects $w$.
     
  3. If $M$ loops on $w$ and $w\in L$, then $M$ does not recognize $L$.
     
  4. If $M$ loops on $w$ and $w\in L$, then $L$ is not recognizable.

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A recognizer $M$ for $L$ satisfies

$w\in L\iff M\ \mathrm{accepts}\ w$.

A is true. If $M$ rejects $w$, it did not accept $w$. 

$\therefore w\notin L$

B is false. For a nonmember, a recognizer is allowed to either $\boxed{\mathrm{reject}}$ or $\boxed{\mathrm{loop\ forever}}$.

So $w\notin L$ does not force rejection.

C is true. If $w\in L$, a recognizer for $L$ must eventually accept $w$. If $M$ loops on that member, $M$ cannot be a recognizer for $L$.

D is false. It only proves that this particular machine $M$ does not recognize $L$. Some other TM could still recognize $L$.

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