Suppose we wish to prove $P$ implies $Q$. We assume $P$. Proof by contradiction begins by assuming not $Q$, and from these two assumptions, a "contradiction" is derived. Now, sometimes that contradiction is that $Q$ is true. I can't help but to feel the argument is circular, yet I can't make precise what it is that bothers me. We're saying not $Q$ implies $Q$, hence $Q$ must be true to begin with. Why are proofs by contradiction of these kind rigorous?
Logic – Understanding Proof by Contradiction and Circular Reasoning
logic
Related Solutions
Proof of negation and proof by contradiction are equivalent in classical logic. However there are not equivalent in constructive logic.
One would usually define $\neg \phi$ as $\phi \rightarrow \perp$, where $\perp$ stands for contradiction / absurdity / falsum. Then the proof of negation is nothing more than an instance of "implication introduction":
If $B$ follows from $A$, then $A\rightarrow B$. So in particular: If $\perp$ follows from $\phi$, then $\phi \rightarrow \perp$ ($\neg \phi$).
The following rule is of course just a special case:
If $\perp$ follows from $\neg \phi$, then $\neg \neg \phi$.
But the rule $\neg \neg \phi \rightarrow \phi$ is not valid in constructive logic in general. It is equivalent to the law of excluded middle ($\phi\vee \neg \phi$). If you add this rule to your logic, you get classical logic.
Your point 5 is correct: given that $P \land Q$ eventually leads to a contradiction, you can conclude that $P \land Q$ is not true.
However, that does not invalidate any of the inferences you made from the assumption that $P \land Q$. That is, from the assumption that $P \land Q$, you can still infer $P$ as well as $Q$, and given those, you can respectively infer $R$ as well as $\neg R$ is true. And hence, you can infer a contradiction. All those inferences are still correct. And hence you can still say that "If $P \land Q$ is true, then $P$, $Q$, $R$, $\neg R$, and $\bot$ (logic symbol for contradiction) are all true as well". But, of course, since a contradiction can never be true, it follows that $P \land Q$ cannot be true either.
OK, so far so good. You also said that you were trying to prove $(P \land Q) \to R$. And, once again, you assumed $P \land Q$, inferred $P$, and from $P$ you were able to infer $R$. At that point, you can indeed infer $(P \land Q) \to R$. Once again, you have basically shown that "If $P \land Q$ is true, then $R$ is true as well", and by conditional proof we can wrap that up with $(P \land Q) \to R$.
Now (and I think this is your real/main question): does the fact that you can also infer $Q$, and from that $\neg R$, and thus a contradiction, take anything away from that? No. It is still true that "If $P \land Q$ is true, then $R$ is true as well". So, when you do a conditional proof, there is really no need to see if your assumption leads to a contradiction.
In fact, if some assumption does lead to a contradiction, then anything follows from your assumption, since anything follows from a contradiction. Hence the conditional that has the assumption as its antecedent ('if' part) will definitely be the case, no matter what you put down for the consequent ('then' part)
Best Answer
The following is a tautology: $$(R\to \neg R) \to \neg R.$$ That is, if from a premise we can prove the premise is false, then the premise is false.
If from $P$ and $\neg Q$ we can prove $Q$, then from $P$ we can prove $\neg Q\to Q$; replacing $R$ with $\neg Q$ in the tautology above, we obtain $$(\neg Q\to \neg(\neg Q))\to \neg\neg Q.$$
In
non-intuitionistic logic,classical logic, the Law of the Excluded Middle tells us that $\neg\neg Q\leftrightarrow Q$ is a tautology. By substituting and using Modus Ponens, we then have that from $P$ we can deduce $ Q$. Since from $P$ we can deduce $Q$, we conclude that $P\to Q$ holds.That is:
$$\begin{align*} 1.&&P,\neg Q&\vdash Q&\text{(what we are assuming we can prove)}\\ 2.&&P&\vdash \neg Q\to Q &\text{(Deduction Theorem)}\\ 3.&&P&\vdash ((\neg Q\to \neg\neg Q)\to \neg\neg Q) &\text{(Tautology)}\\ 4.&&P&\vdash \neg\neg Q\leftrightarrow Q &\text{(Tautology)}\\ 5.&&P&\vdash (\neg Q\to Q)\to Q &\text{(by substitution of 4 in 3)}\\ 6.&&P&\vdash Q &\text{(Modus ponens, 2,5)}\\ 7.&&&\vdash P\to Q &\text{(Deduction Theorem)} \end{align*}$$
That said:
It is very common to find what I call "fake proofs by contradiction." Proofs in which we wish to prove $P\to Q$; we assume both $P$ and $\neg Q$. Then without using the assumption $\neg Q$, we proceed to use the assumption $P$ to establish $Q$. Then the author says "But this contradicts $\neg Q$, a contradiction; therefore, $Q$."
That is bad form and a badly written proof. The proof can be done directly by simply removing the line that says "Assume $\neg Q$", and removing the line that says "But this contradicts $\neg Q$, a contradiction." That is, what we have is a direct proof, that has been turned into a proof by contradiction by adding an extra assumption at the beginning (which is never used) and an extra line at the end (which only notes that the new extra line at the beginning contradicts our earlier conclusion).
For example:
Fake proof by contradiction. Assume to the contrary that $V=W_1\cup W_2$.
If $W_1\subseteq W_2$, then $W_1\cup W_2=W_2\neq V$, because we are assuming $W_1$ and $W_2$ are proper subspaces. So we may assume that $W_1\not\subseteq W_2$. Symmetrically, if $W_2\subseteq W_1$, then $W_1\cup W_2=W_1\neq V$, so we may assume $W_2\not\subseteq W_1$.
Let $w_1\in W_1-W_2$, and $w_2\in W_2-W_1$. Then $w_1+w_2\notin W_1$ (since $w_1\in W_1$ but $w_2\notin W_1$); and $w_1+w_2\notin W_2$ (since $w_2\in W_2$ but $w_1\notin W_2$). Hence $w_1+w_2\notin W_1\cup W_2$. Thus, $w_1+w_2\in V-(W_1\cup W_2)$, so $V\neq W_1\cup W_2$.
But this contradicts our assumption that $V=W_1\cup W_2$. Therefore, this assumption is false, and so we conclude that $V\neq W_1\cup W_2$. $\Box$
Now notice that if we delete the first lines ("Assume to the contrary...") and the last two lines ("But this contradicts our asumption that..."), we have a perfectly fine direct proof of what we were trying to prove! The extra lines do nothing but obscure the proof and possibly create confusion. So it's better to take them out.
Perhaps that is the kind of thing you observed, since you specifically mention a "proof by contradiction" in which the contradiction is arrived at by showing that $P$ and $\neg Q$ imply $Q$?
These are valid proofs, they just include stuff that is not needed.
Note that you are not assuming $Q$, you are assuming $\neg Q$; since you are not assuming what you want to prove, you are not engaging in circular reasoning.