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 $x\neq y$:
$$\exists x \exists y[(R(x) \land R(y) \land x\neq y]$$
A complement to the comments above.
It's worth mentioning that the meaning of the propositional connectives $\neg$, $\wedge$, $\vee$, $\to$ should not be regarded as a mere symbolic translation of the meaning of their English counterparts "not", "and", "or", "if ... then" respectively. See Hedman's A First Course in Logic (2004), p.1-2:
Unlike their English counterparts, these symbols represent concepts that are precise and invariable. The meaning of an English word, on the other hand, always depends on the context. For example, ∧ represents a concept that is similar but not identical to “and.” For atomic formulas A and B, A ∧ B always means the same as B∧A. This is not always true of the word “and.” The sentence
She became violently sick and she went to the doctor.
does not have the same meaning as
She went to the doctor and she became violently sick.
Likewise ∨ differs from “or.” Conversationally, the use of “A or B” often precludes
the possibility of both A and B. In propositional logic A∨B always means
either A or B or both A and B.
This is the case in the sentences above:
(1) Catch Billy a fish, and you will feed him for a day.
(2) Teach him to fish, and you'll feed him for life.
Note that the "and" here should not be interpreted as "$\wedge$". We have many similar cases:
(i) Jump and you die
We intuitively know that this sentence actually means:
(i') If you jump then you will die
Hence the argument is stated this way:
- $C \to D$
- $T \to L$
$\therefore \neg L \vee T$
Which is not valid (note that the conclusion says '$L \to T$').
Best Answer
first, do you believe the the argument is true? And how did you come to that belief? (that might help later when you do symbolic manipulation)
'using logical equivalences' means replace parts of the sentence with equal parts. e.g. $X \rightarrow Y$ can be replaced by $\neg X \lor Y$
the [kinds of equivalences you might use here...modus tollens: replace $X\rightarrow Y$ with $\neg Y \rightarrow \lnot X$ (that's a true equivalence, right?) and $W \land (W \lor Z)$ with $W$. Repeat until you get what you want.
For example of modus tollens, if as part of a larger statement, you can replace $(X\rightarrow Y) \land \neg Y$ with $\neg X$ because they are equivalent (because given that $X$ implies $Y$, if you also know that $Y$ is false then you can infer that $X$ cannot be true, so $\neg X$ is true).