# Thread: Determine whether he following logic argument is valid.

2. ## Re: Need help in solving this question

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.

3. ## Re: Need help in solving this question

How do you get (9) from 8? Truth table gives F T F F which is something different.

4. ## Re: Need help in solving this question

Originally Posted by terrorsquid
How do you get (9) from 8? Truth table gives F T F F which is something different.
$\displaystyle \frac{\alpha \land \beta}{\alpha }$

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 $\displaystyle \alpha \land \beta \models \beta$

5. ## Re: Need help in solving this question

Originally Posted by Also sprach Zarathustra
$\displaystyle \frac{\alpha \land \beta}{\alpha }$

and,

$\displaystyle \frac{\alpha \land \beta}{\beta }$
Is there somewhere I can read more about that? is it a law of logic? I don't understand.

6. ## Re: Need help in solving this question

Originally Posted by Also sprach Zarathustra
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 $\displaystyle \alpha \land \beta \models \beta$
How do we know that r is true and that s is false though? such that $\displaystyle (r\wedge -s) \equiv~$ $\displaystyle (T\wedge -F)$

Not following sry :S

7. ## Re: Need help in solving this question

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.

8. ## Re: Need help in solving this question

A word about not and LaTeX.
$$\neg P$$ gives $\displaystyle \neg P$