Conjunction Elimination and Implication (Is a test for contradiction necessary?)

Question

I am interested in whether it is required to test for contradictions in assumptions within proofs, together with the rules of working with conjunction elimination & implication (I believe these two questions to be related due to the following). Specifically, consider the following:

1. Assume P ∧ Q
2. Therefore, by conjunction elimination we have P is true and Q is true
3. Lets try to prove P ∧ Q ⇒ R
4. If we can prove P ⇒ R, if my understanding is correct, we have proven P ∧ Q ⇒ R since P ∧ Q ⇒ P ⇒ R. Please correct me if I'm wrong here.
5. But what if we also have Q ⇒ ¬ R? This would create a contradiction and therefore P ∧ Q is false?

As I understand, in proof by contradiction of P ⇒ Q , we would assume ¬ Q and try to prove a contradiction to show that if P is true, then Q must be true. However, here, we are not contradicting any of the original assumptions (P).

The question is what if the original assumption (which is assumed to be true) has a contradiction in it? Does that mean my above logic of using conjugation elimination and then implication is not formally correct and I also have to prove there is no contradiction within any of the assumptions that are combined via conjugation whenever I am trying to show (P ∧ Q) ⇒ R (i.e. I cannot rely on P ⇒ R to show (P ∧ Q) ⇒ R)).

I know I cannot rely on P ⇒ R to show P ∨ Q ⇒ R. What are the ground rules for working with conjugation elimination & implication?

A good summary of the logical implication rules I have found in https://www2.seas.gwu.edu/~ayoussef/cs1311/Logic.pdf (page 5). However I cannot find a good resource on the internet for my query above on the mechanics of working conjunction elimination and then implication.

I believe this answer helps quite a bit in terms of removing the need to test for contradiction within the assumptions: In a proof by contradiction, how do we know the assumption is the cause of the contradiction?

However, I still feel uncomfortable not testing for contradiction when applying conjunction elimination and then implication. If my assumptions like in the above result in two contradicting statements, how do I know for certain which one to choose (and how do I get comfortable that there are no contradicting statements without testing for it). The reason I ask this is because conjunction elimination results in a less general condition (i.e. P is less general than P ∧ Q); I am much more comfortable if I derived my implications directly from the more general case P ∧ Q and accordingly questioning the validity of the approach of relying on P ⇒ R.

Answer 1

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)

Answer 2

$P\to R,Q\to\lnot R$ do entail $P\land Q\to R$.

These premises do also entail $P\land Q\to\lnot R$

This is not by itself a contradiction, however $P\land Q$ would entail a contradiction under those premises.

So these premises entail that $\lnot(P\land Q)$