Translate the following argument into symbolic notation.

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

1 Answer

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 $x\neq y$:

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