Q:

Urgent Help Given p ⇒ q, use the Fitch System to prove ¬q ⇒ ¬p.

I think I'm close but I don't know what to do next. Help, please.
All I've done is this:
1.p=>q  Premise
2.~q    assumption
3.p     assumption
4.q     implication elimination 1,3
5.q&~q  and introduction 2,4
I tried with "negation introduction" 3,5 but I can't.
Here is the link of the exercise.
http://intrologic.stanford.edu/exercises/exercise_04_10.html

A:

We need to prove of $p \to q \vdash \sim q \to p$. 
As our goal has $\to$ as its main logical connective, we will use Conditional Introduction. In your system, has the following format:
$$\phi \vdash \psi \\ \overline{\phi \to \psi}$$
Now, we have to provide a proof of $\sim q \vdash \sim p$. To accomplish it, assume $\sim q$ and derive $\sim p$.
$
\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\def\Ae#1{\qquad\mathbf{\forall E} \: #1 \\}
\def\Ai#1{\qquad\mathbf{\forall I} \: #1 \\}
\def\Ee#1{\qquad\mathbf{\exists E} \: #1 \\}
\def\Ei#1{\qquad\mathbf{\exists I} \: #1 \\}
\def\R#1{\qquad\mathbf{R} \: #1 \\}
\def\ci#1{\qquad\mathbf{\land I} \: #1 \\}
\def\ce#1{\qquad\mathbf{\land E} \: #1 \\}
\def\ii#1{\qquad\mathbf{\to I} \: #1 \\}
\def\ie#1{\qquad\mathbf{\to E} \: #1 \\}
\def\be#1{\qquad\mathbf{\leftrightarrow E} \: #1 \\}
\def\bi#1{\qquad\mathbf{\leftrightarrow I} \: #1 \\}
\def\qi#1{\qquad\mathbf{=I}\\}
\def\qe#1{\qquad\mathbf{=E} \: #1 \\}
\def\ne#1{\qquad\mathbf{\neg E} \: #1 \\}
\def\ni#1{\qquad\mathbf{\sim I} \: #1 \\}
\def\IP#1{\qquad\mathbf{IP} \: #1 \\}
\def\x#1{\qquad\mathbf{X} \: #1 \\}
\def\DNE#1{\qquad\mathbf{DNE} \: #1 \\}
$
$$
\fitch{p \to q}{
 \fitch{\sim q}{
 \vdots\\
    \sim p
}\\
\sim q \to \sim p
}\\
$$
In order to derive $\sim p$, we would need to use Negation Introduction. In your system, has the following format:
$$\phi \to \psi \\ \underline{\phi \to \sim\psi} \\ \sim \phi$$
So, in your example, we need provide to proofs: $p \to q$ and $p \to \sim q$. We already have $p \to q$ as one of the premises, we would set-up the proof of $p \to \sim q$:
$$
\fitch{p \to q}{
 \fitch{\sim q}{
 \fitch{p}{
 \vdots\\
    \sim q
}\\
    \sim p
}\\
\sim q \to \sim p
}\\
$$
In full:
$$
\fitch{1.\, p \to q}{
 \fitch{2.\, \sim q}{
 \fitch{3.\, p}{
 \fitch{4.\, q}{
 5.\, q \R{4}
}\\
6.\, q \to q \ii{4-5}
\fitch{7.\, q}{
 8.\, \sim q \R{2}
}\\
9.\, q \to \sim q \ii{7-8}
10.\, \sim q \ni{6,9}
}\\
11.\, p \to \sim q \ii{3,10}
12.\, \sim p \ni{1,11}
}\\
13.\, \sim q \to \sim p \ii{2,12}
}\\
$$

