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Yes, $NP=P=co-P=co-NP$ Where $P=co-P$ since $P$ is closed under complement (the TMs are deterministic and always halt, simply swap "reject" with "accept") …

Yes: we know that being in $NP$ is equivalent to having a polynomial verifier. Let $M$ be the verifier of $B$ (getting an input $x$ and a witness $w$), and let $f$ be the reduction function. … M(f(x),w) \iff f(x)\in B\iff x\in A$, and thus $M'$ is a verifier for $A$, hence $A\in NP$ …

As you have stated, $Composites\in NP$ and $\overline{Primes}=Composites$. Hence, what you proved is that $Primes\in co-NP$. …

The problem with your proof is the fact that a TM for $NP$ would be non-deterministic. … If not, try to think of $NP$ as polynomial verifiable languages, i.e languages with a machine $M(x,w)$ where $w$ is a "witness" describing the $\exists$, and then calling a routine for an NP problem means …

There is a distinction between $S\in (NP\cup co-NP)$, and $S\in (NP\lor co-NP):=\{A\cup B|A\in NP, B\in co-NP\}$ You assumed both are the same, however the first talks about $S$ being in either $NP$ or … If $NP\neq co-NP$, then there is some $L\in NP\setminus co-NP$ (or the other way around), which means that $L\in (NP\cup co-NP)$ but $\overline{L}\notin (NP\cap co-NP)$ (since $L\notin NP$ then $\overline …

Being in $NP-complete$ is equivalent to being in both classes: NP NP-hard In particular, $B\in NP$. … Now, $A\in NP$ since $A\le_p B$ and $NP$ is closed under poly-time reductions (you can easily see how to create a non-deterministic poly-time TM for $A$ given the reduction $f$ and a TM for $B$) …

Take a look at the Cook Levin theorem, that shows a reduction from any $NP$ problem to $SAT$. Since $Dominating-Set\in NP$, this is a reduction $Dominating-Set\le_p SAT$. …

The verification algorithm will just use the solving algorithm and will ignore the verification "proof" it gets. I.e, a verifier is a TM $V(x,w)$, such that $x\in L\iff \exists w, |w|\text{ is polynom …

Double the graph size: make two clones of the input $G_1,G_2$ and now create the (not connected) graph $\hat G$ that will consist of the two clones $G_1,G_2$. Now a half-hamiltonian path in $\hat G$ i …

The second approach is technically the correct interperation. When we say "a verifier gets a certificate $c$ that looks like [...] and computes something", then we actually mean that this verifier che …

What would happen if $P=NP$? Clearly $P$-complete $=P$, and hence also $NP$-complete = $NP$. This means that every language in $NP$ would also be in $NP$-complete. … So finding a language that isn't $NP$-complete, means that we must have $P\neq NP$ …

An $n\times n$ grid takes $n^2$ memory cells. You don't need to remember the "history" of the game, as it is irrelevant.

$P$ is closed under complement. The rest is up to you.

Proving $NP=co-NP$ doesn't necessarily mean that $P=NP$. Although, the other way around is correct: Assume $P=NP$, then $co-NP=co-P=P=NP$. …

Take a look at the definition of NP-complete, it might help. How fast can we compute $MST^*$? Try to use this in conjunction with the assumption you made. …