Why is the assumption needed in this conditional introduction? In the first derivation detailed here, why must we include a subderivation with $P$ as an assumption? We can derive $Q$ (4) from $S \land Q$ (2) without the help of $P$ (3); and then since we have shown that the consequent is true, we should be able to attach any arbitrary antecedent without having to assume that it too is true, since $P \rightarrow Q$ is true for any $P$ when $Q$ is true.
 A: It depends on the formal system you work with.
From the looks of it, the formal system you've referred to has no axioms (which some people think problematic).  Thus, there doesn't exist any way to pass from Q to (P→Q), without first assuming "P" and then deriving "Q".
However, if you have ($\alpha$→($\beta$→$\alpha$)) as an axiom, then from Q as true by one application of detachment (and making appropriate substitutions) you can pass to (P→Q) immediately in one step.  ($\alpha$→($\beta$→$\alpha$)) is also a sensible axiom to have around for a natural deduction system with axioms, since along with ((($\alpha$→($\beta$→$\gamma$))→(($\alpha$→$\beta$)→($\alpha$→$\gamma$))), detachment and uniform substitution, you can prove the Deduction (meta) Theorem and thus conditional introduction becomes a derivable rule of inference.
More simply also, if you have ($\alpha$→($\beta$→$\alpha$)) and ((($\alpha$→($\beta$→$\gamma$))→(($\alpha$→$\beta$)→($\alpha$→$\gamma$))) as axioms you can prove that ((($\alpha$→$\beta$)→$\gamma$)→($\beta$→$\gamma$)) (which appears as an axiom in some bases of propositional calculus).  So, given Q and ((P→Q)→R) you can pass to R in two detachments.
