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If you give a word $w$ not in the language, then the TM is not guaranteed to halt as an NPDA isn't always guaranteed to terminate in finite steps for a finite word.
So, you can produce a counter example in which your construction doesn't work. Take an NPDA which doesn't terminate for a given word (exists, of course) and run your algorithm on that word. ...
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Original Question:
$L_1 = \{w \in \{0|1\}^* | \text{ w is a sequence of one or more 1's } \}$
$L_2 = \{\langle M \rangle | \text{ Turing machine } M \text{ decides } L \}$
prove that $L_2$ is undecidable.
Answer:
For first part you are correct.
First language $L_1$ is indeed decidable. Actually it is regular as represented by regular expression $1^+$.
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