I have the following predicate:
P(x,y) = x is better than y
and was given the following quantified statement for the sentence "exactly one person is better than Sam":
$$\exists x [P(x, Sam) \land (\neg \exists y P(y, Sam)\land (y \neq x))]$$
I am wondering if this is logically equivalent to:
$$\exists x \exists y[(P(x,Sam) \land P(y,Sam)) \rightarrow (x = y)]$$
Intuitively I can't see why they are not, but am having a difficult time converting one to the other via logical laws.
Best Answer
No. $\neg \exists y \phi \not \equiv \exists y \neg \phi$, so you can not apply the negation just to the conjunction and move the existential quantifier out. What is the negated is the existence of other better persons. Note also that the $\exists y$ is supposed to scope over the whole second conjunction; it's $P(y,Sam) \land (y \neq x)$ which belongs together, but in your second sentence you changed the bracketing structure so as to $P(x,Sam)$ to belong to the first part and applying the negation just to $y \neq x$.
The second sentence can be satisfied in a structure where noone is better than Sam. Because then for arbitrary choices of $x$ and $y$ the conjunction becomes false so the implication becomes true so the existential statement becomes true, which is not the intended meaning.
The first sentence entails the second, but not vice versa, so they are not equivalent.
A logically equivalent transformation via logical laws is
$\begin{align*} & \exists x [P(x, Sam) \land \neg \exists y [P(y, Sam) \land (y \neq x))]\\ \equiv\ & \exists x [P(x, Sam) \land \forall y \neg [P(y, Sam)\land (y \neq x)]]\\ \equiv\ & \exists x [P(x, Sam) \land \forall y [P(y, Sam) \to (y = x)]]\\ \end{align*}$