Use predicate logic to express the statement "No one has more than three cousins" using the propositional function $C(x, y)$ , which represents "$x$ is the cousin of $y$".
Answer:
$C(x,y) \ $ represents the statement $x$ is the cousin of $y$
Then , No one has more than three cousin becomes
$$\forall x \ \forall y \ (C(x,y) \to x \leq 3 ) $$
But I am not sure.
Help me out
Let's start with the negative because it is simpler:
Exists someone with more than 3 cousins, so $(\exists x)(\exists y,z,w,p)((y\ne z\land y\ne w\land y\ne p\land z\ne w\land x\ne p\land w\ne p)\land(C(y,x)\land C(z,x)\land C(w,x)\land C(p,x)))$
In words: exists $x$ such that exists 4 different people, $y,z,w,p$, such that they all are cousins of $x$.
Now changing it to negative will be: for all $x$ and for all different $4$ people, $y,z,w,p$, at least one of them is not a cousin of $x$:
$(\forall x)(\forall y,z,w,p)((y\ne z\land y\ne w\land y\ne p\land z\ne w\land x\ne p\land w\ne p)\rightarrow (\lnot C(y,x)\lor\lnot C(z,x)\lor\lnot C(w,x)\lor\lnot C(p,x)))$.
You can do:
$$\forall x \exists y_1 \exists y_2 \exists y_3 \forall z (C(z,x) \rightarrow (z = y_1 \lor z = y_2 \lor z = y_3))$$
This works, because for any $x$ it rules out any fourth possible cousin. The use of the existentials makes it look like it is asserting the existence of some cousins, but note the statement is also true if no one has any cousins at all, because then $C(z,x)$ will simply be false for any $z$.
The nice thing about this method is that you don't have to write down a long list of non-identity statements.
Hint: $\forall x\forall y_1\forall y_2~((C(y_1,x)\wedge C(y_2,x))\to y_1=y_2)$ says "no one has more than one cousin".
