Without constructing a truth table show that the statement formula ~(~p→~q)→~(q→p) is a tautology I've been trying to solve this for a few days now but I'm not sure if I'm right. I wasn't able to attend some important lecturers so I'm not sure if this is how I should present the final answer or if the final answer is right.
After going through an old post in this site I tried to establish a base like this;

*

*p = It is raining

*q = there are clouds

*~p = It is not raining

*~q = there are no clouds

So starting from left to right we have;
~(It is not raining implies there are no clouds) also implies ~(There are clouds implies it is raining).
Negating the first part gives (It is raining implies there are clouds).
Negating the second part gives (There are no clouds implies it is not raining).
This sign ($\rightarrow$) between these two parts is what confuses me, logically both statements when joined makes sense, is that what a tautology means? If so how should I best present this as an answer, and if not how do I go about solving it?
I'm not sure about the tags I should use on this post, I'm new here.
 A: Remember that $p \vee \neg p$ is a tautology, meaning that the statement always evaluates to true. Also, $p \rightarrow q \equiv \neg p \vee q$.
So we have
\begin{align}
\neg (\neg p \rightarrow \neg q) \rightarrow \neg (q \rightarrow p) &\equiv \neg \left[\neg(\neg p) \vee \neg q\right] \rightarrow \neg (\neg q \vee p) & \text{(definition of implication (x2) )}\\
&\equiv \neg \left(p \vee \neg q\right) \rightarrow \neg (\neg q \vee p) \\
&\equiv \neg \left(p \vee \neg q\right) \rightarrow \neg (p \vee \neg q) & \text{(rearranging terms)}\\
&\equiv \neg \left[\neg (p \vee \neg q) \right] \vee \neg (p \vee \neg q) & \text{(definition of implication))} \\
&\equiv (p \vee \neg q) \vee \neg ( p \vee \neg q)
\end{align}
Now let $r = p \vee \neg q$. We then have $(p \vee \neg q) \vee \neg (p \vee \neg q) \equiv r \vee \neg r$, which is a tautology.
A: Here's a proof using natural deductions rules:

*

*$\underline{\mid\quad} ¬(¬P→¬Q)~$ - assumption

*$\mid\quad\underline{\mid\quad} Q→P~$ - assumption

*$\mid\quad\mid\quad\underline{\mid\quad} ¬P~$ - assumption

*$\mid\quad\mid\quad\mid\quad\underline{\mid\quad} Q~$ - assumption

*$\mid\quad\mid\quad\mid\quad\mid\quad P~$ - E→ in 2,4

*$\mid\quad\mid\quad\mid\quad\mid\quad ⊥~$ - in 3,5

*$\mid\quad\mid\quad\mid\quad ¬Q~$ - I¬ in 4 (close assumption 4)

*$\mid\quad\mid\quad ¬P→¬Q~$ - I→ in 3,7 (close assumption 3)

*$\mid\quad\mid\quad ⊥~$ - in 1,8

*$\mid\quad ¬(Q→P)~$ - I¬ in 2 (close assumption 2)

*$¬(¬P→¬Q)→¬(Q→P)~$ - I→ in 1,10 (close assumption 1)

