Translate from English to first-order logic: There is somebody who likes everyone who likes John

Question

Translate from English to first-order logic: There is somebody who likes everyone who likes John

I'm not sure about the answer. What I'm thinking is $(∃x)(L(x,y)∧L(x,J))$

Answer 1

$$(\exists x)(\forall y)(L(y,J)\rightarrow L(x,y))$$

Translation:

There exists $x$ such that if $y$ likes John, then $x$ likes $y$.

Answer 2

The English language is not logically precise, so the sentence has two (possible more) interpretations. The first, as given by @quasi, is "there is somebody who likes (everybody who likes John)", and has the translation $(\exists x)(\forall y)(L(y,J)\rightarrow L(x,y))$. The second is "there is (somebody who likes everybody) who likes John", which has the translation $(\exists x)[(\forall y(L(x,y)))\,\land\,L(x,J)]$. Here $L(a,b)$ means $a$ likes $b$.

(The first is probably the intended meaning, because if somebody liked everyone, they would surely like John. I'm just pointing out an interpretation closer to what you were originally thinking.)

Answer 3

There is some person who likes everyone who likes John. So, there is some person $x$ who likes all people $y$ who satisfy the condition that $y$ likes John. So, there exists a person $x$ such that for all $y$, if $y$ likes John then $x$ likes $y$. We have the expression: $\exists x, \forall y (L(y,J) \implies L(x,y))$.