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Translate the following argument into symbolic notation.

Use $P(x)$ for "$x$ is purple",
$Q(x,y)$ for "$x$ questions $y$" and
$R(x)$ for "$x$ is ridiculous".

Somebody who is purple questions somebody.
Somebody who is not purple questions somebody.
Everybody who questions anybody is ridiculous.
Therefore, at least two people are ridiculous.

I have so far done the following; however, I am not sure if I am correct. Once I find the notation to the sentences I have to find a proof which I understand how to. But before that I need to make sure I have the symbolic notation correct. Can someone please help check where I it’s wrong, or a hint will do as well.

$\exists x (P(x) \wedge Q(x,y))$

$\exists x (\neg P(x) \wedge Q(x,y))$

$\forall x \bigl(P(x)\implies \exists y(R(x))\ \bigr)$

Therefore, $\exists x (R(x))$

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    $\begingroup$ Hints: your final argument should have no free variables. On the LHS there is an unbound y, and the translation of 'everybody who questions anybody is ridiculous' is completely off (it actually uses the 'purple' predicate and not the 'questions' predicate.) $\endgroup$
    – Steven Stadnicki
    Dec 5, 2012 at 0:54

1 Answer 1

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The first two expressions are a start, but you need to quantify $y$, where $y$ the somebody who is questioned!

  • Somebody who is purple questions somebody.

$$\exists x \exists y[P(x) \land Q(x, y)]$$

  • Somebody who is not purple questions somebody.

$$\exists x\exists y[\lnot P(x) \land Q(x, y)]$$

Third expression:

  • Everybody who questions anybody is ridiculous. (Or: "everyone who questions any someone is ridiculous.") I think you want to use the predicate $Q(x,y)$ here:

$$\forall x[\exists yQ(x, y) \rightarrow R(x)]$$

Conclusion:

  • "Therefore at least two people are ridiculous." Here you need to declare the existence of at least two distinct people $x$ and $y$, so you need to include $x\neq y$:

$$\exists x \exists y[(R(x) \land R(y) \land x\neq y]$$

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  • $\begingroup$ The first two expressions are still unbound in $y$ and need to be corrected for that. $\endgroup$
    – Steven Stadnicki
    Dec 5, 2012 at 1:28
  • $\begingroup$ @ Steven, they were being edited while you commented. $\endgroup$
    – amWhy
    Dec 5, 2012 at 5:23
  • $\begingroup$ Saba Di: have you figured the proof using the logical translations? You reach the same result even if you used the original first two expressions. We just needed to ensure that the $y$ in the first two expressions was quantified. $\endgroup$
    – amWhy
    Dec 5, 2012 at 5:24

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