Logic is something I am entirely self-taught in. Due to my resulting ignorance, it is difficult for me to search for the right thing. Therefore, please excuse me if this has been asked before.
The Details:
In this YouTube video by Attic Philosophy, an approach to doing natural deduction proofs is demonstrated by neither proven nor rigorously defined; it is just an introduction.
Essentially, one works backwards. The example given is a proof of
$$p\to (q\to r)\vdash (s\to p)\to(q\to(s\to r)).$$
We start by assuming $p\to (q\to r)$, leaving a big gap, then writing $(s\to p)\to(q\to(s\to r))$, like so:
$$\begin{array}{|lc} p\to (q\to r) & \\ \hline & \\ & \\ & \\ & \\ & \\ & \\ & \\ (s\to p)\to (q\to (s\to r)). & \end{array}$$
Then we assume the first bit of the conclusion and do the same thing as above, like so:
$$\begin{array}{|lc} p\to (q\to r) & \\ \hline \rlap{\begin{array}{|lc} s\to p & \\ \hline & \\ & \\ & \\ & \\ & \\ q\to (s\to r) & \end{array}}&\\ (s\to p)\to (q\to (s\to r)). & \end{array}$$
And again:
$$\begin{array}{|lc} p\to (q\to r) & \\ \hline \rlap{\begin{array}{|lc} s\to p & \\ \hline \rlap{\begin{array}{|lc} q & \\ \hline & \\ & \\ & \\ s\to r & \\ \end{array}}& \\ q\to (s\to r) & \end{array}}&\\ (s\to p)\to (q\to (s\to r)). & \end{array}$$
Once more:
$$\begin{array}{|lc} p\to (q\to r) & \\ \hline \rlap{\begin{array}{|lc} s\to p & \\ \hline \rlap{\begin{array}{|lc} q & \\ \hline \rlap{\begin{array}{|lc} s & \\ \hline & \\ & \\ r \end{array}}& \\ s\to r & \\ \end{array}}& \\ q\to (s\to r) & \end{array}}&\\ (s\to p)\to (q\to (s\to r)). & \end{array}$$
Now we fill in the gap using all the assumptions we like from above it, like so:$\hskip 36em {\require{cancel}\require{cancelto} _\text{psst! over here!}\cancelto{\hspace{1pt}}{\hspace{20pt}}}$
