# Need help solving this argument....PLEASE

• Oct 21st 2007, 07:28 AM
ambergarrett955
1. A v (B -> C)
2. B / A v C
3.
• Oct 21st 2007, 09:01 AM
Plato
Quote:

Originally Posted by ambergarrett955
1. A v (B -> C)
2. B
3./ A v C

Do by indirect argument.
Assume ~(AvC) then ~A ^ ~C.
~A by simplification.
(B->C) by disjunctive syllogism.
~C by simplification.
~B by modus tollens.
• Oct 21st 2007, 09:12 AM
ambergarrett955
I'm trying to get A v C by itselfs, though. So, is that still the answer? Thank you
• Oct 21st 2007, 09:52 AM
Plato
Quote:

Originally Posted by ambergarrett955
I'm trying to get A v C by itselfs, though. So, is that still the answer?

The answer is yes only if you know how to use indirect argument.
If you do not have that tool, then I do not how to help you with this.
• Oct 21st 2007, 11:00 AM
ambergarrett955
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 AM
Soroban
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: .$\displaystyle A \vee C$