Categorical proof in natural deduction I'm reading Fitch's book on Symbolical Logic and I don't understand how to prove, with natural deduction, that the following is a theorem without using any hypothesis. This is what is to be proven (categorically): $$ [[p\to q]\to [[q \to r]\to [s\to [[s\to p]\to r]]]]$$
 A: Hint
1) $p \to q$ --- assumed [a]
2) $q \to r$ --- assumed [b]
3) $s$ --- assumed [c]
4) $s \to p$ --- assumed [d]
With three consecutive applications of $\to$-elim yu can derive $r$.
The "re-assemble" the formula with $\to$-intro.
A: A conditional proof consists of assuming the antecedent aiming to derive the consequent, thereby allowing conditional introduction.  
Now, you seek to prove a nested conditional statement of the form $A\to(B\to (C\to (D\to E)))$.  $$(p{\to}q)\to\big((q{\to}r)\to\big(s\to\big((s{\to}p)\to r\big)\big)\big)$$
Therefore your proof should be built as a nest of conditional proofs, assuming $A,B,C,D$ aiming to derive $E$, then introducing the four conditionals in turn. 
The only source of confusion might be that three of the antecedents you need to assume are themselves conditionals.  Don't worry, just do it. 
Assume $p\to q, q\to r, s, s\to p$, aiming to derive $r$ followed by four conditional introductions to discharge each of the assumptions.
That derivation will involve conditional eliminations, because, frankly, what else are you going to do with those assumed conditional statements? 
