Do I have to discharge an antecedent that I assume?

Question

For example, if I have the premise:

$P \rightarrow (Q \rightarrow R)$

Can I assume P to get:

$Q\rightarrow R$

And then assume Q to get R.

For reductio ad absurdum and arrow introduction I know that you have to discharge the assumptions that you use, I was just wondering if this is the case for assuming the antecedent.

Here is an argument that seems to require the assumption of the antecedents.

$E\rightarrow (\sim F \lor \sim (A \lor D)) \therefore E \land F \rightarrow \sim D \lor G$

Answer 1

For example, if I have the premise:

$P \rightarrow (Q \rightarrow R)$

Can I assume P to get:

$Q\rightarrow R$

And then assume Q to get R.

You may do that, but discharging those assumptions gets you right back to $P\to(Q\to R)$

Suppose however that you assumed $Q$ then assumed $P$, to derive $R$. Discharging the assumptions will deduce $Q\to(P\to R)$ is entailed by the premise.

But $R$ is not entailed by $P\to(Q\to R)$, alone.

$R$ is entailed by $P\to (Q\to R)$, $P$, and $Q$.

Here is an argument that seems to require the assumption of the antecedents.

$E\to(\lnot F \lor \lnot (A \lor D)) \therefore E \land F \to \lnot D \lor G$

This is how a Conditional Proof works in Natural Deduction.   Here you assume $E\land F$, derive $\lnot D\lor G$, then discharge the assumption via conditional introduction.

$$\def\fitch#1#2{\quad\begin{array}{|l} #1\\\hline #2\end{array}}\fitch{E\to (\neg F\lor\lnot (A\lor D))\quad:\textsf{Premise}}{\fitch{E\land F\quad:\textsf{Assumption}}{F\quad:\textsf{Conjunction Elimination}\\E\quad:\textsf{Conjunction Elimination}\\\lnot F\lor\lnot(A\land D)\quad:\textsf{Conditional Elimination}\\\quad\vdots\\\lnot F\to \lnot D\quad:\textsf{Conditional Introduction}\\\quad\vdots\\\lnot(A\lor D)\to\lnot D\quad:\textsf{Conditional Introduction}\\\lnot D\quad:\textsf{Disjunction Elimination}\\\lnot D\lor G\quad:\textsf{Disjunction Introduction}}\\(E\land F)\to(\lnot D\lor G)\quad:\textsf{Conditional Introduction}}$$

Answer 2

You could do so, but the proof would be unfinished. You could discharge each premise in turn to get:

$$P\implies [Q \implies R] \implies [P\implies [Q \implies R]]$$

If you introduced all of your premises at once, you could also prove:

$$[P\implies [Q \implies R] \land P \land Q] \implies R$$