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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?

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    $\begingroup$ There's a free variable $x$ that doesn't make much sense. Perhaps the last predicate should be $\neg L(s,c_3)?$ $\endgroup$ – saulspatz Mar 4 '18 at 22:52
  • $\begingroup$ yes, that's what I meant, my bad for the typo $\endgroup$ – waterbottle Mar 4 '18 at 23:05
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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 $c_3$.


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 $\forall c_3 \in CM,\ c_1 \ne c_3 \wedge c_2 \ne c_3 \Rightarrow \cdots$.

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  • $\begingroup$ That makes a lot of sense. I'm just wondering what the difference would be then if we were to say ∃c1 ∈ CM, ∃c2 ∈ CM, c1≠c2? What really is the difference between "There exist two distinct cabinet ministers" and "For any two distinct cabinet ministers"? $\endgroup$ – waterbottle Mar 5 '18 at 4:50
  • $\begingroup$ Suppose there are 100 ministers. If you use $\forall$, then it says that no matter which two you pick, there is some movie star who likes them (and only them). On the other hand, if you use $\exists$, then it says that there are two ministers who have a movie star who like them (and only them) but there could very well be 98 ministers who are hated by all movie stars. $\endgroup$ – Clive Newstead Mar 5 '18 at 4:57
  • $\begingroup$ In translating "∀c1 ∈ CM, ∀s ∈ MS, ~(∃c2 ∈ CM,c1≠c2 ^ L(c2,s)) -> ~ L(c1,s)" would I be right in saying "for any cabinet minister c1, if there does not exist a distinct cabinet minister that likes all movie stars then no cabinet minister likes any cabinet ministers." So could the interpretation be that "all cabinet ministers dislike all movie stars if there isn't' another cabinet minister that likes all movie stars?" $\endgroup$ – waterbottle Mar 5 '18 at 5:18
  • $\begingroup$ @simba: No. When you say "[if] there does not exist a [disctinct] cabinet minister that likes all movie stars", you're saying $\neg \exists c_2 \in CM,\ \forall s \in S, L(c_1,s)$, which is not what you have. Note that the implication part of that statement is of the form $\neg P \Rightarrow \neg Q$, which is the contrapositive of $Q \Rightarrow P$. As such, it is equivalent to saying: "If any minister likes a movie star, then there is another minister that likes the same movie star." $\endgroup$ – Clive Newstead Mar 5 '18 at 5:30
  • $\begingroup$ my mistake, I should've noticed right away the position of the negation. Thank you so much for your explanation! $\endgroup$ – waterbottle Mar 5 '18 at 5:40

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