We have to deduce $\displaystyle t\rightarrow w$, which is equivalent to $\displaystyle \neg w \rightarrow \neg t$, from:
$\displaystyle \\ \neg p \rightarrow (r \wedge \neg s) \quad (1) \\ t \rightarrow s \quad (2) \\ u \rightarrow \neg p \quad (3) \\ \neg w \quad (4) \\ u \vee w \quad (5) \\ u \quad (6) \mbox{ From 5 and 4} \\ \neg p \quad (7) \mbox{ From 3 and 6 } \\ r \wedge \neg s \quad (8) \mbox{ From 1 and 7 } \\ \neg s \quad (9)\mbox{ From 8 } \\ \neg t \quad (10) \mbox{ From 2 and 9 } \\ \neg w \rightarrow \neg t \quad \mbox{ From 4 and 10 }$
Hence, it's a valid argument. Hope this helps you.
$\displaystyle \frac{\alpha \land \beta}{\alpha } $
and,
$\displaystyle \frac{\alpha \land \beta}{\beta }$
Edit:
Proof:
$\displaystyle \alpha \land \beta$ true only when $\displaystyle \alpha$ and $\displaystyle \beta $ are both true. Therefor, if $\displaystyle \alpha \land \beta$ true at particular reference, so are $\displaystyle \alpha$ and $\displaystyle \beta$ are true in that particular reference.
And we can write, $\displaystyle \alpha \land \beta \models \alpha$ and $\displaystyle \alpha \land \beta \models \beta$
We have correctly inferred that $\displaystyle r \wedge \sim s$ is true, but this implies that $\displaystyle r$ is true and so is $\displaystyle \sim s$.
Simplification - Wikipedia, the free encyclopedia
I'm sorry I couldn't find something else.