## Proof: is the language $L_1$$={|$$\emptyset \subseteq L(M)$} (un)-decidable?

I want to show that $$L_1 =$${$$| \emptyset \subseteq L(M)$$} is decidable/undecidable – without rice theorem (just for the case that I can apply it).

Every language contain the $$\emptyset$$ as a subset. So my guess is that the language is decidable.

Therefore, let us assume that $$L_1$$ is decidable. Lets say that $$N$$ is the TM which decides $$L_1$$.

N = "with input $$$$:"

How can I proof that $$N$$ is a decider for $$L_1$$?

## $L = \{ \langle M \rangle \mid M$ is a turing machine and $L(M)$ is decidable $\}$

$$L = \langle M \rangle \mid M$$ is a turing machine and $$L(M)$$ is decidable $$\}$$. How could I show this language is undecidable using the HALTing method?

A language L is undecidable if its not corecognizable and recognizable.

So I wish to show $$HALT \leq_m \overline{L}$$ for not recognizable and

$$HALT \leq_m L$$

Not corecognizable

SHOWING Not recognizable attempt:

R = "On input <M, w>     Construct M1     M1 = "On input x          run M on w          accept     return <M1> 

If $$\langle M, w \rangle \in HALT$$, then M1 accepts all input $$L(M_1) = \Sigma^*$$ which is undecidable $$\langle M_1 \rangle \notin L$$ as we would want

If $$\langle M, w \rangle \notin HALT$$, then M1 loops on all input $$L(M_1) = \emptyset$$, which is decidable $$\langle M_1 \rangle \in L$$ as we would want

After further look into this $$L(M_1) = \Sigma^*$$ is actually decidable? How would I do this then. What can I do to M1 to make it return an undecidable TM if M halts on w?

## prove $L=\{M\ |\ M\ is\ a\ TM\ and\ \forall.x\in \Sigma^*\ with\ |x|>2,\ M\ on\ x\ runs\ at\ most\ 4|x|^2\ steps\}\notin R$

I am trying to prove that the language

$$L=\{M\ |\ M\ is\ a\ TM\ and\ \forall.x\in \Sigma^*\ with\ |x|>2,\ M\ on\ x\ runs\ at\ most\ 4|x|^2\ steps\}$$

belongs to $$Co-RE$$ but not to $$R$$.

Showing $$\bar{L}\in RE$$ is pretty much straight forward, but I also want to show that $$L\notin{R}$$

My idea was a reduction $$\bar{H_{TM}}\le_mL$$ but I struggle to figure out how to do it.

Any help/guidance will be much appreciated.

## L: $\{\langle M \rangle \mid M$ is a TM and $\{111\} \subseteq L(M)\}$ basically 111 can get accepted by $L(M)$

L: $$\{\langle M \rangle \mid M$$ is a TM and $$\{111\} \subseteq L(M)\}$$ basically 111 can get accepted by $$L(M)$$

Proof that this is recognizable:

Here is a recognizer $$M1$$ for L:

Let M1 = "On input ⟨M⟩: 1. Run M on 111 2. if M accepts then accept else loop" 

Is this right? if $$\langle M \rangle \in L$$ then it will accept $$\langle M \rangle \notin L$$ then it will loop. I’m attempting to do this without dove tailing but cant find a lot of resources for this

## Complexity of the language of all TMs $M$ such that $L(M)$ is decidable

Let  R = \{\langle M \rangle \mid L(M) \text{ is decidable}\}. Is $R$ recursively enumerable or co-recursively enumerable?