Some boys like every girl

Question

How to write this into predicate logic ?

Some boys like every girl

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I wrote two predicate logic sentences:

$\forall(x)(girl(x) \rightarrow \exists(y) (boy(y) \wedge likes(y,x)))$

and

$\exists(x)(boy(x) \wedge \forall(y)(girl(y) \rightarrow likes(x,y)))$

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Obviously, the second one is right. But I am confused for 1st one. I guess its also right.

As I think English translation of 1st one - "Every girl is liked by some boy"

Answer 1

As I see it, there are two parts to this question: understanding exactly what we mean by "some boys like every girl," and writing down symbols that capture that meaning. I think "some boys like every girl" is a little ambiguous. It could mean either:

1. There is at least one boy with the property of liking every girl.
2. There is more than one boy with the property of liking every girl.
3. Every girl has the property of being liked by at least one boy.

I personally read "some boys like every girl" as either the first or the second, depending on how much one reads into the fact that "boys" is plural. I can certainly see people reading it any of these three ways, though; it's frustratingly ambiguous. Translating directly into propositional logic notation, these interpreations are:

1. \begin{align*} \exists (x) \operatorname{boy}(x) \wedge \left( \forall (y) \operatorname{girl}(y) \rightarrow \operatorname{likes}(x,y) \right) \end{align*}
2. \begin{align*} \exists (x_0) \exists (x_1) & \left( x_0 \neq x_1 \right) \wedge \\ & \left( \operatorname{boy}(x_0) \wedge \left( \forall (y) \operatorname{girl}(y) \rightarrow \operatorname{likes}(x_0,y) \right) \right) \wedge \\ &\left( \operatorname{boy}(x_1) \wedge \left( \forall (y) \operatorname{girl}(y) \rightarrow \operatorname{likes}(x_1,y) \right) \right) \end{align*} (This says that there are two $x$'s—which we call $x_0$ and $x_1$—each with the property described above, and that they can't be the same.)
3. \begin{align*} \forall (y) \operatorname{girl}(y) \rightarrow \left( \exists (x) \operatorname{boy}(x) \wedge \operatorname{likes}(x,y) \right)\end{align*}

There are two main things to pay attention to:

• Pay attention to the order of quantifiers. Which one is "outside" and which one is "inside?" In particular "for all $x$ there is a $y$" means there can be a different $y$ associated to each $x$, while "there is a $y$ such that for all $x$" means that the same $y$ will have to do for every $x$.
• Pay attention to how you talk about what "universe" you're quantifying over. "All boys have property $P$" should be "$\forall (x) \operatorname{boy}(x) \rightarrow P(x)$", but "there is a boy with property $P$" should be "$\exists (x) \operatorname{boy}(x) \wedge P(x)$".

Your first attempt is exactly the same as what I wrote for interpretation 3, and you analyzed it correctly. Your second attempt is almost what I wrote for interpretation 1, except that you make an error in how you specified the "universes" for your quantifiers. What you wrote literally translates to "there is something that, if it that thing is a boy, everything is a girl liked by the first thing," which is pretty clearly not what you wanted ;-).

Answer 2

$\newcommand{\boy}{\operatorname{boy}}\newcommand{\girl}{\operatorname{girl}}\newcommand{\likes}{\operatorname{likes}}$ $\forall (x)(\girl(x) \to\exists (y) (\boy(y) \land \likes(y,x)))$

"Every girl is liked-by some boy."   Wilma is liked by Fred, Betty by Barny, ...

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$\exists(x)(\boy(x) \color{crimson}\land \forall(y)(\girl(y) \color{crimson}\to \likes(x,y)))\qquad$ (edit: note the correction.)

"There is some boy who likes every girl."   His name is Bill.

Answer 3

Presumably your quantifiers range over all the boys and girls. The second is incorrect because the forall would require that all the children are girls. The second will be true only if all the children are girls, in which case the antecedent is false and the implication is true. The fix is simple. Instead of and you should have implies, which would say that for all the children, if they are a girl then $x$ likes them, which is what you want. Then you want your implies to be and. As written, it is true if there is one girl, because then there is an $x$ such that the antecedent is false so the (old) implication is true.

Your first says that for each girl there is some boy that likes her. It is not necessary that the same boy like every girl.