How to say in FOL:
- "francis" is "Pope"
- Nothing is identical to "francis"
- Except for "francis", is false that something is "Pope"
The formula should imply that:
- $P(f)$ is TRUE
- $P(f)$ is NOT IDENTICAL to $P(g), P(h), P(i)$ etc.
- $P(g), P(h), P(i)$ etc. are FALSE
I think this is not satisfying:
$$P(f)∧∀x(P(x)→f=x)$$
because it doesn't imply that $P(g), P(h)$ etc. are false, only that they are identical to $P(f)$.
In FOL, if you want to express there is exactly one element in the given domain with property $P$, then we write
$$ \exists y [P(y) \wedge \forall x [P(x) \to y=x] ] $$
Now, if you want to identify the unique element with property $P$ as Francis, abbreviated as $f$, then we let $y=f$ and apply existential instantation to obtain
$$ P(f) \wedge \forall x [P(x) \to f=x] ] $$
which does in fact imply that $P(g), P(h)$, etc. are false statements. To see how, let $x=g$ where $g$ is an individual constant corresponding to someone who is not Francis, say Gary. In other words, let $x=g$ where $f \neq g$. Now, assume the statement $P(g)$ is true. Then, by Modus Ponens we obtain $f=g$, implying Francis and Gary are the same person. But we have already stated $f \neq g$. Hence, we have a contradiction. Since we obtain a contradiction under the assumption $P(g)$ is true, it must be the case $P(g)$ is false.
Note the above formula can be written equivalently as
$$ P(f) \wedge \forall x [f \neq x \to \neg P(x)] ] $$