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Let M represent the set of all Mathematics courses and S represent the set of all students.

Predicates:

E(s, m) : "student s is enrolled in math course m"

C(s): "student s owns a cat"

I want to transform the two sentences below using quantified predicates.

  1. There is at least one math student who owns a cat. (A math student is a student who is registered in at least one math class)

  2. There is at least one student in every math class who owns a cat.

I got this:

1) ∃s∈M, E(s, m)-> C(s)

or perhaps

∃E(s,m), C(s)

Is this correct?

2) ∃s∈M, ∀m∈M, C(s) ?

Also for:

∀s∈S, ∀m∈M, [E(s,m) -> ¬C(s)]

does this translate to:

"All student cat owners are not registered in any math courses"

Thank you.

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  • $\begingroup$ For the first, I would prefer $\exists s \exists m(E(s,m)\land C(s))$. For the second, $\forall m\exists s(E(s,m)\land C(s))$. Your version of (2) is not correct, it says it is the same student in all the classes who has a cat. $\endgroup$ – André Nicolas Oct 3 '13 at 0:08
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For (1), the formula "$\exists s \in M$, $E(s, m)\to C(s)$" translates to "there is a math course $s$ such that if $s$ is enrolled in $m$, then $s$ has a cat". This is not what you wanted to say. And the formula $\exists E(s,m), C(s)$ makes no sense because quantifiers can only be applied to variable symbols, not predicate symbols.

To write "there is at least one math student who owns a cat," where a math student is a student who is enrolled in at least one math class, we need to write "there is at least one student who is enrolled in at least one math class and who owns a cat." This can be done as $\exists s\,\exists m, (E(s,m) \wedge C(s))$ as André says, or alternatively as $\exists s, ((\exists m, E(s,m)) \wedge C(s))$ because "$\exists m, E(s,m)$" says that $s$ is a math student.

For (2), the formula "$\exists s\in M, \forall m \in M, C(s)$" says "there is a math course $s$ such that for all math courses $m$, $s$ owns a cat. Again not what you wanted to say: for one thing, courses cannot own cats.

For your third question ("Also for:") the statement "$\forall s \in S, \forall m \in M, (E(s,m) \to \neg C(s))"$ translates to "for all students $s$ and math courses $m$, if $s$ is enrolled in $m$, then $s$ does not have a cat." This is equivalent to "no math student has a cat," which in turn is equivalent to what you wrote.

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  • $\begingroup$ If courses can have cats, then sign me up! $\endgroup$ – Trevor Wilson Oct 3 '13 at 0:38
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For the first question, use something like $$\exists s \exists m(E(s,m)\land C(s)).$$ If you want to specify membership, that's fine, but one should be consistent. So you could write $$(\exists s\in S)( \exists m\in M)(E(s,m)\land C(s)).$$

For the second, $$\forall m\exists s(E(s,m)\land C(s))$$ will work.

Your version of (2) is not correct. Since you put the existential quantifier in front, it says it says in particular that there is a particular student who is enrolled in every single math class. Very busy.

On the third question, the translation is correct. Perhaps it could be made a little more idiomatic. Something like "No math student owns a cat."

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  • $\begingroup$ For ∃s∃m(E(s,m)∧C(s)), do you mean ∃s∈m(E(s,m)∧C(s))? So that there exists some student in a math course... $\endgroup$ – muros Oct 3 '13 at 0:23
  • $\begingroup$ We can add the belonging stuff. I added to my answer to give that version. But you can't say $\exists s\in M$, since $M$ is a set of courses, not a set of students. $\endgroup$ – André Nicolas Oct 3 '13 at 0:30
  • $\begingroup$ I'm not sure what you mean about (2). What the OP wrote for (2) didn't even mention the $E$ (enrollment) predicate. $\endgroup$ – Trevor Wilson Oct 3 '13 at 0:39
  • $\begingroup$ Also both of the OP's answers for (1) are wrong, no? You say "I would prefer..." but your verdict is not clear. $\endgroup$ – Trevor Wilson Oct 3 '13 at 0:41
  • $\begingroup$ I have changed it. Maybe I wanted to be nice. The suggested answers for the first question basically made no sense. (Unlike the second question, where the answer was clearly enough expressed that one could call it wrong.) $\endgroup$ – André Nicolas Oct 3 '13 at 1:05

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