Translate from English to first-order logic: There is somebody who likes everyone who likes John 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 $(∃x)(L(x,y)∧L(x,J))$
 A: $$(\exists x)(\forall y)(L(y,J)\rightarrow L(x,y))$$
Translation:
There exists $x$ such that if $y$ likes John, then $x$ likes $y$.
A: 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.)
A: 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: $\exists x, \forall y (L(y,J) \implies L(x,y))$. 
