My question is regarding Example 2.2.2. of Velleman's 'How to Prove It' (pg. 71). We are asked to write the statement "John likes exactly one person" in terms of quantifiers (note, $\exists!$ has not been introduced yet).
My attempt:
I started by breaking the statement down as follows:
"There exists a person $x$ that John likes and John does not like all other persons $y$ who are not $x$."
and then:
"There exists a person $x$ that John likes and for all persons $y$, if $y \neq x$, then John does not like them."
and then expressed this using quantifiers:
$$∃x[L(j, x) ∧ ∀y[(y ≠ x) → ￢L(j, y)]]$$
where $L(j, m)$ is the statement "John likes m."
Questions
- Is it correct to use an implication here?
I know that in order for my answer to be correct, the statement $(y ≠ x) → ￢L(j, y)$ must be true for all $y$. But I'm not sure if it is in this case, since John likes exactly one person, so it it is not possible for ￢L(j, y) to be true if (y ≠ x) is false.
- Would a biconditional be better? i.e. $∃x[L(j, x) ∧ ∀y[(y ≠ x) ↔ ￢L(j, y)]]$.
The solution given by the author is $∃x(L( j, x)∧¬∃y(L( j, y) ∧ y \neq x))$. I tried checking if my solution was equivalent, as follows:
$$\begin{align*} ¬∃y[L(j, y) ∧ y ≠ x] \\& = ∀y¬[L(j, y) ∧ y ≠ x] &&\text{(quantifier negation)} \\& = ∀y[¬L(j, y) ∨ ¬y ≠ x] &&\text{(DeMorgan)} \\& = ∀y[¬L(j, y) ∨ y = x] &&\text{(double negation)} \\& = ∀y[L(j, y) → y = x] &&\text{(conditional)} \\& = ∀y[y ≠ x → ¬L(j, y)] &&\text{(contrapositive)} \end{align*}$$
So it seems my solution is equivalent, but I'm still not sure why it's ok to use an implication to express "exactly one".
If you use a biconditional, then, translating back into English, you statement would mean:
There exists a person $x$ that John likes and for all persons $y$, if $y \ne x$, then John does not like $y$, and if $y = x$ then John does like $y$.
Now, to say "for all persons $y$, if $y = x$ then John does like $y$" is just a roundabout way of saying "John likes $x$." So by changing the conditional to a biconditional, you are, in effect, adding "and John likes $x$" to your statement. But you've already said that. So using a biconditional isn't wrong, but it's redundant.