I'm reading our lecturers recitations, and the lecturer remarked some comment that confuses me.
Consider the expression $\exists !x:P(x)$, where $P$ is some predicate.
This expresses the existence of a unique value for which the predicate $P$ holds.
In logical form, this can also be written as: $\exists x:(P(x)\wedge \neg\exists y:(P(y)\wedge y\ne x))$, and I can see why, but then it says there in a small remark: "Observe the difference between that and $(\exists x:P(x))\wedge (\neg\exists y:(P(y)\wedge y\ne x))$. can you see why the latter form is not valid?"
I tried to think about it but couldn't figure out why is it not valid.
Can someone please clarify this for me?
In the first expression :
I've slightly modified the parentheses in order to highlight the scope of the first existential quantifier: the whole formula, with the two occurrences of the variable $x$ bounded by $∃x$.
The fact that both occurrences are in the scope of the quantifier means that they are referring to the same object.
In the other expression :
the first existential quantifier bound only the variable $x$ of $P(x)$.
The occurrence of $x$ in the subformula $(¬∃y:(P(y)∧y≠x))$ is free; this means that it can refer to an objcet different from the one for which $P$ holds of, according to the sub-formula $(∃x:P(x))$.
Consider as example the domain $\mathbb N$ of natural numbers and consider as interpreation of the predicate $P(x)$ the relation : $(x=0)$.
With this interpretation the first formula means :
which is a (trivial but) true statement regarding the natural numbers and it express exactly the fact taht $0$ is the unique number satisfying the formula $(x=0)$.
If we "instantiate" the existential quantifier with $x:=0$ we get :
that is clearly true ; we cannot find a value for $y$ such that $(y=0)∧(y≠0)$ holds.
Consider now the second expression; with the above interpretation it means :
The left-concjunct is clearly true ; it's enough to choose $0$ as value for the variable $x$.
But now, the right-conjunct has a free occurrence of $x$ and we are not "forced" to assign to it the same value used in the left-conjunct.
We can choose $1$ as value for the free $x$ in the right-conjunct and what we get is :
But now the formula is false; it is enough to choose $0$ also as value for $y$ and we have that $(0=0)∧(0≠1)$ holds. Thus, it is not true that $¬∃y:((y=0)∧y≠1)$.