0
$\begingroup$

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".

$\endgroup$
5
  • $\begingroup$ 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 $\lnot a$. $\endgroup$ Commented Oct 14, 2012 at 10:04
  • $\begingroup$ Step 2 should read $\neg a \wedge \neg x$ by De Morgan's laws $\endgroup$ Commented Oct 14, 2012 at 10:04
  • $\begingroup$ Corrected, but how did step 2 go to step 3. $\endgroup$
    – Samuel
    Commented Oct 14, 2012 at 10:09
  • $\begingroup$ $p\land q\to p$, 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 $\lnot x$. $\endgroup$ Commented Oct 14, 2012 at 10:12
  • $\begingroup$ every step means having a $\rightarrow$ in the middle. I thought is the "=" sign.=.=' OIC, thanks. $\endgroup$
    – Samuel
    Commented Oct 14, 2012 at 10:20

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .