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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.

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  • $\begingroup$ "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." Yes, and how do you propose this is done in a formal proof? $\endgroup$ – Git Gud Aug 6 '14 at 2:03
  • $\begingroup$ I don't understand your question. Given $Q$, should $P \rightarrow Q$ not immediately follow, whatever $P$ may be? Assuming $P$ to be true seems redundant. (But thank you for correcting my tags.) $\endgroup$ – Malnormalulo Aug 6 '14 at 2:13
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    $\begingroup$ I think you're confusing semantics with syntax.It's hard for me to explain exactly what I think your confusion is without pretty much typing everything there is to say. I'll try to focus on the essential part. Natural deduction is a game. You have some rules (the rules of inference) and you need to use them. In the linked question $P\to Q$ doesn't mean "$P$ implies $Q$". The intuition behind is that, that is correct. But this is purely syntactic. You should look at $P\to Q$ simply as a string of symbols. You can even replace all instances of $\to$ with $\spadesuit$. $\endgroup$ – Git Gud Aug 6 '14 at 2:20
  • $\begingroup$ Ah, I think I see. So $P \rightarrow Q$ results not from the mere fact that its consequent is affirmed, but from the fact that $Q$ is valid given a set of statements consisting of the premises and the antecedent; and it is this set and only this set that demonstrates that the material conditional is valid, so it is irrelevant whether the antecedent actually contributed to proving the consequent's validity. $\endgroup$ – Malnormalulo Aug 6 '14 at 2:39
  • $\begingroup$ "Only this set that demonstrates that the material conditional is valid" Not sure what set you mean, but there are infinite different ways to infer $P\to Q$ from those premises. Despite of this, you seem to have the right idea. $\endgroup$ – Git Gud Aug 6 '14 at 2:43
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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.

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