1. A v (B->C)

2. B/ A v C

3.

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- Oct 21st 2007, 07:28 AMambergarrett955Need help solving this argument....PLEASE
1. A v (B

**->**C)

__2. B__/ A v C

3. - Oct 21st 2007, 09:01 AMPlato
- Oct 21st 2007, 09:12 AMambergarrett955Reply to answer
I'm trying to get A v C by itselfs, though. So, is that still the answer? Thank you

- Oct 21st 2007, 09:52 AMPlato
- Oct 21st 2007, 11:00 AMambergarrett955
In class we always came out with the answer that was behind this /xx as our last #. I'm just confused. This stuff is so confusing to me. I'm sorry.

- Oct 21st 2007, 11:30 AMSoroban
Hello, Amber!

Quote:

$\displaystyle \begin{array}{c}A \vee (B \to C)\\

B\qquad\qquad\quad \\ \hline

A \vee C\qquad\quad\end{array}$

I vaguely recall a theorem . . . the Law of Detachment (?)

. . $\displaystyle \begin{array}{c} p \vee q \\ \sim q\quad \\ \hline p\quad \end{array}$

The first line is: .$\displaystyle A \vee (B \to C) \quad\Rightarrow\quad A \vee (\sim\!B \vee C)\quad\Rightarrow\quad (A \vee C)\: \vee \sim\!B$

So the argument becomes: .$\displaystyle \begin{array}{c}(A \vee C)\:\vee\: \sim\!B \\ B\qquad\qquad \\ \hline \end{array}$

By the Law of Detachment, the conclusion is: .$\displaystyle A \vee C$