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Convert the following into formal predicate logic. Define predicates as necessary. Then negate the predicate sentence. Push all negations to the closest terms.

1) There are at least two people who everyone knows. Domain = {People}

2) Every student takes at least two classes. Domain = {people, classes}

3) All Students know each other. Domain ={ All people}

My take in this.... for the 1) part is it valid to do something like this ∃𝒙∃𝒚∀z𝑷(𝒙, 𝒚,z). Where in my words i could be totally wrong.. there exist a pair (x,y) who everyone (z) knows.

For the 2) part i know that there has to be and existential quantifier and a universal. but I'm not sure how to write them down.

For the Last part 3) is it this ∀𝒙∀𝒚𝑷(𝒙, 𝒚) just on the basis that for all X there is a Y

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  • $\begingroup$ People knowing each other should be expressed as a binary relation, not a ternary relation, so your solution to $1$ isn't right. Also, you need to expressly note (in formal logic) that $x$ and $y$ are not the same person. $\exists x, y \forall z P(x, z) \land P(x, y)$ is not a correct solution because it's possible that $x$ and $y$ are the same person. Your solution to $3$ isn't quite correct because your domain is all people, but your statement is only supposed to be true of all students. So you need a relation symbol to denote that a member of your domain is a student. $\endgroup$ – Robert Shore Sep 18 '19 at 23:35
  • $\begingroup$ Could you help with the final solution? For the 3rd statement do you suggest using p(x) and q(x) is that what you mean? @RobertShore $\endgroup$ – Elchavo18 Sep 18 '19 at 23:44
  • $\begingroup$ I'd strongly prefer to provide hints. For $3$, use $S(x)$ to denote "$x$ is a student." And a minor tweak to what I wrote for $1$ will give you a correct solution. $\endgroup$ – Robert Shore Sep 18 '19 at 23:46
  • $\begingroup$ Hey the hints help with the learning process. Much appreciated. I just ask for some patience if i ask a lot of questions. Any hint for the second part while i work on the other 2? @RobertShore $\endgroup$ – Elchavo18 Sep 18 '19 at 23:51
  • $\begingroup$ You'll need separate predicates to define "people" and "classes" and a binary relation to express that a person is taking a class. $\endgroup$ – Robert Shore Sep 19 '19 at 0:07
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1) There are at least two people who everyone knows. Domain = {People}

My take in this.... for the 1) part is it valid to do something like this ∃𝒙∃𝒚∀z𝑷(𝒙, 𝒚,z). Where in my words i could be totally wrong.. there exist a pair (x,y) who everyone (z) knows.

You must say: "There are some $x$ and some $y$ who are not the same people and every $z$ will know $x$ and know $y$."

You should also use a bivariate predicate such as $\def\op#1{\operatorname{\rm #1}}\op{K}(~,~)$ for "_ knows _"

$$\exists x~\exists y~\forall z~\bigl(x\neq y\wedge \op{K}(z,x)\wedge \op{K}(z,y)\bigr)$$


2) Every student takes at least two classes. Domain = {people, classes}

"For every $x$ who is a student, then there is an $y$ which is a class that is taken by $x$ and there is a $z$ which is another class that is taken by $x$."

You will need predicates for: $\op P(~)$ "_ is a people", $\op C(~)$ "_ is a class", and $\op T(~,~)$ "_ takes _" .


3) All Students know each other. Domain ={ All people}

"For every $x$ who is a student, then for every $y$ who is a student, then $x$ knows $y$."

Use predicates $\op S(~)$ for "_ is a student", and $\op{K}(~,~)$ for "_ knows _".

(NB: do you need to worry about whether $x$ and $y$ are the same people?)

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  • $\begingroup$ Thank You! Your break downs help a lot. I think I got it from here. $\endgroup$ – Elchavo18 Sep 19 '19 at 1:40

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