Problems in formalizing these sentences This is the first sentence that I have to formalize:

"Every student likes at least one type of cake"

Let:


*

*$S(x)$ stands for 'x is a student'

*$C(x)$ stands for 'x is a type of cake'

*$L(x,y)$ stands for 'x likes y'


My formalisation is:
$\forall x (S(x)\rightarrow \exists y (C(y) \wedge L(x,y)))$
Is it correct? 
Or should I write:
$\forall x \exists y (S(x)\wedge C(y) \rightarrow L(x,y))$ ?
The second sentence is:

"There is a type of cake such that every student likes it"

Is the following formalisation correct?
$\exists x (C(x) \wedge \forall y (S(y) \rightarrow L(y,x)))$
Many thanks for the help
 A: In short:


*

*formalization $\forall x (S(x)\rightarrow \exists y (C(y) \wedge L(x,y)))$ is correct;

*formalization $\forall x \exists y (S(x)\wedge C(y) \rightarrow L(x,y))$ is not correct, suppose that there is an object $y_0$ which is not a cake (perhaps a student), then setting $y = y_0$ would make the formula true even if all students dislike all cakes;

*formalization $\exists x (C(x) \wedge \forall y (S(y) \rightarrow L(y,x)))$ is correct.


To give you some intuition:


*

*You could think of unary predicates as sets, in particular


$$\forall x \in S\ (\Phi)\quad \text{ is equivalent to } \quad\forall x \big(S(x) \to \Phi\big)$$
$$\exists x \in S\ (\Phi)\quad \text{ is equivalent to } \quad\exists x \big(S(x) \land \Phi\big)$$


*

*Your first formula is $\forall x \in S\ \Big( \exists y \in C\ \big(L(x,y)\big)\Big)$.

*Your second formula can be rewritten as
\begin{align}
&\forall x \exists y \big(S(x)\wedge C(y) \rightarrow L(x,y)\big)\\
&\forall x \exists y \Big(S(x)\to \big(C(y) \to L(x,y)\big)\Big)\\
&\forall x \Big(S(x)\to \exists y \big(C(y) \to L(x,y)\big)\Big)\\
&\forall x \in S\ \exists y \Big(C(y)\land\big(C(y) \to L(x,y)\big) \lor \neg C(y)\land\big(C(y) \to L(x,y)\big)\Big)\\
&\forall x \in S\ \exists y \Big(C(y)\land L(x,y) \lor \neg C(y)\Big)\\
&\forall x \in S\ \Big(\exists y \big(C(y)\land L(x,y)\big) \lor \exists y \big(\neg C(y)\big)\Big)\\
&\forall x \in S\ \Big(\exists y \in C\ \big(L(x,y)\big) \lor \exists y \in \overline{C}\Big)\\
\end{align}
which could be read as "for every student there is a cake that he likes or something which is not a cake".

*Your third formula is $\exists x \in C\ \Big(\forall y \in S\ \big(L(y,x)\big)\Big)$.


I hope this helps $\ddot\smile$
