Determine whether he following logic argument is valid.

**MathsNewbie0811** · August 21st 2011, 12:16 AM

I am stuck in this question. Please help.

$Determine whether he following logic argument is valid.-untitled.jpg$

**MATHNEM** · August 21st 2011, 12:50 AM

We have to deduce $t\rightarrow w$ , which is equivalent to $\neg w \rightarrow \neg t$ , from:

$\\ \neg p \rightarrow (r \wedge \neg s) \quad (1) \\ t \rightarrow s \quad (2) \\ u \rightarrow \neg p \quad (3) \\ \neg w \quad (4) \\ u \vee w \quad (5) \\ u \quad (6) \mbox{ From 5 and 4} \\ \neg p \quad (7) \mbox{ From 3 and 6 } \\ r \wedge \neg s \quad (8) \mbox{ From 1 and 7 } \\ \neg s \quad (9)\mbox{ From 8 } \\ \neg t \quad (10) \mbox{ From 2 and 9 } \\ \neg w \rightarrow \neg t \quad \mbox{ From 4 and 10 }$

Hence, it's a valid argument. Hope this helps you.

**terrorsquid** · August 21st 2011, 01:36 AM

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

**Also sprach Zarathustra** · August 21st 2011, 01:53 AM

Originally Posted by terrorsquid

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

$\frac{\alpha \land \beta}{\alpha }$

and,

$\frac{\alpha \land \beta}{\beta }$

Edit:

Proof:

$\alpha \land \beta$ true only when $\alpha$ and $\beta$ are both true. Therefor, if $\alpha \land \beta$ true at particular reference, so are $\alpha$ and $\beta$ are true in that particular reference.

And we can write, $\alpha \land \beta \models \alpha$ and $\alpha \land \beta \models \beta$

**terrorsquid** · August 21st 2011, 02:03 AM

Originally Posted by Also sprach Zarathustra

$\frac{\alpha \land \beta}{\alpha }$

and,

$\frac{\alpha \land \beta}{\beta }$

Is there somewhere I can read more about that? is it a law of logic? I don't understand.

**terrorsquid** · August 21st 2011, 02:42 AM

Originally Posted by Also sprach Zarathustra

Proof:

$\alpha \land \beta$ true only when $\alpha$ and $\beta$ are both true. Therefor, if $\alpha \land \beta$ true at particular reference, so are $\alpha$ and $\beta$ are true in that particular reference.

And we can write, $\alpha \land \beta \models \alpha$ and $\alpha \land \beta \models \beta$

How do we know that r is true and that s is false though? such that $(r\wedge -s) \equiv~$ $(T\wedge -F)$

Not following sry :S

**MATHNEM** · August 21st 2011, 11:43 AM

We have correctly inferred that $r \wedge \sim s$ is true, but this implies that $r$ is true and so is $\sim s$ .

Simplification - Wikipedia, the free encyclopedia

I'm sorry I couldn't find something else.

**Plato** · August 21st 2011, 11:47 AM

A word about not and LaTeX.
[tex]\neg P[/tex] gives $\neg P$

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