I study for an exam and I have this exercise that I've practiced
Let: Monkey(x) = "x is a monkey"
Likes(x, y) = "x likes y"
∃x (Monkey(x) ∧ ∀y(Monkey(y) ∧ x ̸= y ⇒ Likes(y, x)))
My response:
There exists someone to like all monkeys, but not all monkeys like every one.
I could use some feedback, please.
Let's write
A=Monkey(x)
B=∀y(Monkey(y) ∧ x ̸= y ⇒ Likes(y, x))
We look at "∃x(A and B)". So there exists x such that A and B both holds.
By A, x is a monkey.
By B, we know the following fact: "if y is a monkey and y is not x, then y likes x".
As a conclusion, we obtain that: "there is a monkey (let's call him x) such that every other monkey likes him (i.e. any monkey that is not x likes x) . "