# About proving a arguments is valid

I have a question is about proving a argument is valid or not. Again, cannot really understand the solution.

The question is like this

Determine if the following arguments are valid.

1. It is not the case that IBM or Xerox will take over the copier market. If RCA returns to >the computer market, then IBM will take over the copier market. Hence, RCA will not return >to the computer market.

The solution is like this.

Let a denote “IBM will take over the copier market”, x “Xerox will take the copier market”, r “RCA returns to the computer market”. Then we have the following argument:

$$\lnot(a\lor x)$$ $${r \rightarrow a}\over {so \quad \lnot r }$$

1. $\lnot (a \lor x)$ $\quad$ premise
2. $\lnot a \land \lnot x$ $\quad$ from 1
3. $\lnot a$ $\quad$ from 2
4. $r \rightarrow a$ $\quad$ premise
5. $\lnot a \rightarrow \lnot r$ $\quad$ from 4
6. $\lnot r$ $\quad$ from 3 and 5

and the statment is valid

Why the step 2 can go to step 3? Obviously, "It is not the case that IBM or Xerox will take over the copier market" is not equal to "It is not the case that IBM take over the copier market".

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You copied (2) incorrectly, or there was a typo in your source: it should be $\lnot a\land\lnot x$, derived from (1) by de Morgan’s law. And from that you clearly can get $\lnot a$. – Brian M. Scott Oct 14 '12 at 10:04
Step 2 should read $\neg a \wedge \neg x$ by De Morgan's laws – Shaktal Oct 14 '12 at 10:04
Corrected, but how did step 2 go to step 3. – Samuel Oct 14 '12 at 10:09
$p\land q\to p$, as you can check from the truth table, so if you have $p\land q$, you can infer $p$. Here $p$ is $\lnot a$, and $q$ is $\lnot x$. – Brian M. Scott Oct 14 '12 at 10:12
every step means having a $\rightarrow$ in the middle. I thought is the "=" sign.=.=' OIC, thanks. – Samuel Oct 14 '12 at 10:20

## 1 Answer

The move from line 2 to line 3 is called conjunction elimination. It says 'if I know that (A and B) is true, then I know that A is true', also 'if I know that (A and B) is true, then I know that B is true' - where A and B are well formed formulas.

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