Prove true in natural numbers (Peano Arithmetic) While reviewing old exercise sheets, I have found this question and am having difficulties understanding some of the logic:
Let $\mathbb{N}$(natural numbers) be a model for Peano Arithmetic, that means $\mathbb{N}$$\vDash$PA. Choose which of the following sentences are true in $\mathbb{N}$.
(a) $\lnot$$\exists$x((sx=0)$\to$(0=s0)) $\to$ (($\lnot$$\exists$x(sx=s0))$\to$(0=s0))
(b) (($\lnot$$\exists$x(sx=s0))$\to$(0=s0)) $\to$ $\lnot$$\exists$x((sx=s0)$\to$(0=s0))
(c) (($\exists$x(sx=0))$\to$(0=s0)) $\to$ $\exists$x((sx=s0)$\to$(0=s0)) 

For (a) I would say that it is true in $\mathbb{N}$, because if you set $\color{red}{x=0}$ then you get $\lnot$$\exists$x((s$\color{red}{0}$=0)$\to$(0=s0)) which is true. Then the second part of the implication I would also set x=0 again and in the end that part ends up also being true, so we have True implies True, hence (a) is true in $\mathbb{N}$.
For (b) I would start off by letting x=0 so then I would get (($\lnot$$\exists$x(s0=s0)) , which is false, and I can see that (0=s0) is also false, so False to False, hence first part of the implication is True. 
Now this is is where i have run into some confusion. My professor had solved the second part of the implication by setting $\color{red}{x=s0}$,so we then have, $\lnot$$\exists$x((s$\color{red}{s0}$=s0)$\to$(0=s0)) , and this part would be false, so all together True to False, hence (b) is false in $\mathbb{N}$.
I can not understand the logic behind setting x=s0 in (b). Why is that allowed, when in (a) we didn't do that.
It would be greatly appreciated, if someone could please explain this and also provide some guidance for (c).
 A: For (b) :

$[¬∃x(sx=s0) → (0=s0)] → ¬∃x[(sx=s0)→(0=s0)]$

the "strategy" of you teacher is to prove it false showing that the antecedent is true and the consequent false.
But the RHS conditional is negated; thus we want that both the LHS : $¬∃x(sx=s0) → (0=s0)$ and the un-negated part of the RHS : $∃x((sx=s0)→(0=s0))$ are true.

i) Consider the LHS; it has $0=s0$ in the consequent, and this is always false (it means : $1=0$).
So, we have to find a value for $x$ such that the antecedent of the LHS conditional : $¬∃x(sx=s0)$ is false.
But $¬∃x(sx=s0)$ is false, because we can find some value for $x$ such that $sx=s0$ is true... it is enough to choose $0$ as value for $x$.
Thus, it is true that "there are values for $x$ such that $sx=s0$ holds", and so the negation of this sentence, i.e. $¬∃x(sx=s0)$ is false.

ii) Conider now the RHS; we want that the un-negated formula : $∃x((sx=s0)→(0=s0))$ is true.
So, the question is : can we find a value $n$ for $x$ such that : $(sn=s0)→(0=s0)$ is true ?
Again, we have the false consequent : $(0=s0)$; thus, in order to "verify" the conditional, we need a value $n$ for $x$ such that $(sn=s0)$ is false.
Now it is enough to choose $1$ (as our $n$) as value for $x$ and we have : $2 \ne 1$, i.e. :

$\lnot (ss0=s0)$;

this shows that "there are values for $x$ such that $sx=s0$ does not hold", i.e. $sx=s0$ is false for some value of $x$.
Summing up, we can find a value for $x$ such that $(sx=s0) \rightarrow (0=s0)$ is : $False \rightarrow TRUE$, i.e. $TRUE$, and so the sentence $∃x[(sx=s0) \rightarrow (0=s0)$ is true.


For (c), taking into account that in $\mathbb N$ $(0=s0)$ is false, we can rewrite it as :

$(∃x(sx=0) → FALSE) → ∃x((sx=s0) → FALSE)$.

Now, it is quite simple to find its truth value : $∃x(sx=0)$ is false in $\mathbb N$, because $0$ is not the successor of any number. Thus, we have :

$(FALSE → FALSE) → ∃x((sx=s0) → FALSE)$, i.e.
$TRUE → ∃x((sx=s0) → FALSE)$.

Now, in order to have it true, we need a true consequent, and so we have to check if we can "verify" $∃x((sx=s0) → FALSE)$. Again, the only way to "verify" it is to find a value for $x$ such that $(sx=s0) → FALSE$ is true, i.e.  we have to find a value for $x$ such that $(sx=s0)$ is false.
So, it is enough to choose $1$ and we have a value for $x$ such that $(sx=s0) → FALSE$ is true, i.e. $∃x((sx=s0) → FALSE)$ is true in $\mathbb N$, and also the complete formula is.
