2
$\begingroup$

Does a counterexample exist for the following argument?

If person A is not home, then person B is. But, if A is not home, then B isn’t. So, they are both home.

Translated to logical notation:

1) $\neg A \to B$

2) $\neg A \to \neg B$

3) $\therefore A \land B$

To my understanding, a counterexample is when the premises are true but the conclusion is false. I've equated them as such, but I'm stuck in proving whether or not there is a contradiction, since there are too many cases to deal with (e.g. $A \land B \equiv F$ has 3 cases). How would I find out whether there's a counterexample or not?

1) and 2) seem contradicting already, but they also have 3 cases each.

$\endgroup$
6
  • 1
    $\begingroup$ A can be home and yet B does not need to. The first two statemets effectively say that A must be home, but they imply nothing about B. $\endgroup$ – Parcly Taxel Sep 17 '16 at 7:07
  • $\begingroup$ Does that mean that 1) and 2) are actually one premise, joined by a conjunction? Also how does this affect the argument? The conclusion says that only one of the two correct cases are true (the TT case). $\endgroup$ – JC1 Sep 17 '16 at 7:16
  • $\begingroup$ The conclusion is false, since there is a possibility for A and B where the premises are true yet the conclusion is false. The correct conclusion should be just A. $\endgroup$ – Parcly Taxel Sep 17 '16 at 7:20
  • $\begingroup$ Ah ok. So a counterexample does exist then? $\endgroup$ – JC1 Sep 17 '16 at 7:22
  • $\begingroup$ Yes. ${} {} {}$ $\endgroup$ – Parcly Taxel Sep 17 '16 at 7:23
0
$\begingroup$

To my understanding, a counterexample is when the premises are true but the conclusion is false.

Correct; more correctly, a counterexample invalidates an argument by making all premises true while falsifying the conclusion.

Here the conclusion is false if either of $A$ or $B$ are false.

Both premises are implications, which are held to be false only if the antecedent is true while the consequent is false.

Since the consequent of both implications are contradictory, then both implications can only be true at the same time if their antecedent is false.   Hence if a counterexample is to be found, $A$ must be true.

The way to make the conclusion ($A\wedge B$) false when $A$ is true is to make $B$ false.   And lo and behold: when $A$ is true and $B$ false then both premises are true.   Thus we have found a counterexample.

Therefore the argument $\neg A\to B, \neg A\to\neg B\vdash A\wedge B$ is proven to be invalid, by way of counterexample.

$\endgroup$
0
$\begingroup$

We can say, from (1) and (2), that:

$$\lnot A\to (B\land \lnot B)$$

As $B\land\lnot B=F$, we have:

$$\lnot A\to F$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.