I'm a bit unsure about how to approach this problem. Any help is appreciated.
Given the following: $$\begin{align} (p \wedge t) &\to (r \vee s) \\ q &\to (u \wedge t) \\ u &\to p \\ ¬s& \end{align}$$ Show that $q \to r$.
I'm a bit unsure about how to approach this problem. Any help is appreciated.
Given the following: $$\begin{align} (p \wedge t) &\to (r \vee s) \\ q &\to (u \wedge t) \\ u &\to p \\ ¬s& \end{align}$$ Show that $q \to r$.
What you require is a "conditional proof". Assume $q$, derive $r$, thereby deducing that $q\to r$.
To get you started, (presuming you are familiar with Fitch style proof displays) your proof should look somewhat like this ...:
$\def\fitch#1#2{~~\begin{array}{|l}#1\\\hline #2\end{array}} \fitch{~1.~q\to (u\land t)\\~2.~u\to p\\~3.~(p\land t)\to (r\lor s)\\~4.~\neg s}{\fitch{~5.~q}{~6.~u\wedge t\quad: 1,5,\to\textsf{elimination (modus ponens)}\\~~~\vdots\\12. ~r}\\13.~q\to r}$
Fill in the missing details.