Can I use use simplification (rule of inference) inside a negated expression? The question gives these premises and asks for the conclusion.

*

*$P \land Q$

*$P \implies \neg(Q\land R)$

*$S\implies R$
Also there are choices: a) $\neg S$ b)$\neg P$ c) $\neg R$ d) None of the above
Here is my approach:

*

*$P \land Q$, premise

*$P$, simplification from 1

*$P \implies \neg (Q\land R)$, premise

*$\neg (Q \land R)$, modus ponens from 2 and 3

*$\neg Q \lor \neg R$, De Morgan's law applied on 4

*$S \implies R$, premise

*$\neg S \lor R$, logically equivalent to 6

*$\neg Q \lor \neg S$, resolution from 5 and 7

At this point, should I choose none of the above? or if I continue as


*$\neg (Q\land S)$, from 8

*$\neg (S)$, simplification of $(Q \land S)$ in step 9

My question: Is step 10 true? can I do simplification inside the negation and then apply the negation? and in this case the answer would be (a).
 A: No, you cannot infer $\neg S$ from $\neg (Q \land S)$!
Imagine that $Q$ is False but $S$ is True. Then $\neg (Q \land S)$, which says that $Q$ and $S$ are not both True would be true itself: indeed it is not the case that both $Q$ and $S$ are True.  However, $\neg S$ would be False.  So, inferring $\neg S$ from $\neg (Q \land S)$ is an invalid inference.
In general, when you try to apply Simplification to an embedded statement, sometimes the result does validly follow from the original statement, but sometimes it does not.  For example, $\neg \neg S$ does follow from $\neg \neg (Q \land S)$. Inferring $P \to S$ from $P \to (Q \land S)$ is also valid ... but inferring $S \to P$ from $(Q \land S) \to P$ is not.
Now, you can come up with some rule that lays out the conditions under which you can apply something like Simplification to an embedded conjunction, but it is not straightforward, and it is easy to make mistakes. So, much better is to simply say that inference rules can never be applied to embedded statements, and that is exactly what almost every logical inference system does.
