Translating multiple universal quantifiers in predicate logic

Question

I'm very confused as to how to interpret predicate logic statements with multiple universal quantifiers sharing the same domain. For example, I'm trying to make sense of the following statement:

CM is the domain of all Cabinet Ministers

MS is the domain of all movie stars

L(x,y): person x likes person y

∀c1 ∈ CM, ∀c2 ∈ CM,c1≠c2 -> ∃s ∈ MS, L(s,c1) ^ L(s, c2) ^ (∀c3∈CM, c1≠c3^c2≠c3 -> ~L(s,c3))

I have no idea what this could mean. It seems to me that it's saying all cabinet ministers are either liked or disliked by a movies star. I interpret ∀c1 ∈ CM to be all cabinet ministers but how can there be ∀c1 ∈ CM and ∀c2 ∈ CM? Are all cabinet ministers split in two groups?

Answer 1

Let's read it from left to right.

$\forall c_1 \in CM,\ \forall c_2 \in CM, \dots$

For any two cabinet ministers $c_1,c_2$ ...

$\dots c_1 \ne c_2 \Rightarrow \dots$

... if $c_1 \ne c_2$, then ...

$\dots \exists s \in MS, \dots$

... there is a movie star $s$, such that ...

$\dots L(s,c_1) \wedge L(s,c_2) \wedge \dots$

... $s$ likes $c_1$ and $s$ likes $c_2$, and ...

$\dots \forall c_3 \in CM \dots$

... for all cabinet ministers $c_3$ ...

$\dots c_1 \ne c_3 \wedge c_2 \ne c_3 \Rightarrow \dots$

... if $c_3$ is neither $c_1$ nor $c_2$, then ...

$\dots \neg L(s, c_3)$

... $s$ does not like $c_3$.

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Now if you were to read the 'English' version of these things, it doesn't make much sense. But putting it together further, we can now say something like this:

For any two cabinet ministers $c_1$ and $c_2$, who are not the same minister, there is a movie star $s$ who likes both $c_1$ and $c_2$ and such that, for any minister $c_3$ other than $c_1$ or $c_2$, the movie star $s$ does not like $c_3$.

Even more concisely:

For any two distinct cabinet ministers, there is a movie star who likes both ministers but does not like any other ministers.

The word 'distinct' is packed up in $c_1 \ne c_2$, and the words 'any other' are packed up in $\forall c_3 \in CM,\ c_1 \ne c_3 \wedge c_2 \ne c_3 \Rightarrow \cdots$.