I am given the following premises:
\begin{equation} \sim p \vee q \rightarrow r \\
s \vee \sim q \\
\sim t \\
p \rightarrow t \\
\sim p \wedge r \rightarrow \sim s \end{equation}
I am suppose to derive $$\sim q$$ from the above.
I am able to get $$\sim p$$ from the third and fourth premises, but I am not sure how to apply the argument forms to the other premises. I am unsure on how to proceed, should I break the premises with three terms into a two parts first?
How would I go about apply argument forms to get the conclusion?
Best Answer
\begin{align} \lnot p \vee q \rightarrow r \tag{1}\\ s \vee \lnot q \tag{2} \\ \lnot t \tag{3}\\ p \rightarrow t \tag{4}\\ \lnot p \wedge r \rightarrow \lnot s \tag{5}\end{align}
As you note in your post, from the fourth premise we have $p\to t$, and we have, in the third premise, $\lnot t$. By modus tollens, we derive $\lnot p$.
Now, since we've derived $\lnot p$ from premises (3), (4), we also have $\lnot p \lor q$, by "addition" to $\lnot p$, (also called "or-introduction" which is shorthand for disjunction introduction).
And from $\lnot p \lor q$, together with the premise (1), we have $r$ by modus ponens.
Now, since we already deduced $\lnot p$, and we just deduced $r$, we can use "And-introduction (conjunction-introduction)" to get $\lnot p \land r$.
Given $\lnot p \land r$, and premise (5): $(\lnot p \land r) \to \lnot s$, we have, by modus ponens, $\lnot s$.
But given our premise (2), $s \lor \lnot q,$ together with $\lnot s$, we deduce $\lnot q$, as desired.