If the cat is on the mat then that cat exists I know that according to existential generalization:
$$Q(a)\rightarrow\exists xQ(x)$$
But is there a way to say something like these (which I understand to be malformed):
$$Q(a)\rightarrow\exists x$$
Or:
$$Q(a)\rightarrow\exists a$$
I'm trying to make an argument of the form "if the cat is on the mat then that cat exists", which I assume is a valid inference (else is it possible that the cat is on the mat but there isn't that cat?).
 A: In first-order logic (with identity for point 2):

*

*The natural way to talk about specific individuals (such as that cat) is to use constant symbols, such as $a$ in the OP. Such constant symbols cannot be quantified, exactly because they refer to specific individual in the domain of discourse.


*The natural way to express an existence predicate $E(y)$ (claiming that "$y$ exists") is the formula $\exists x (x = y)$. See here and here as references for deep  and technical discussions.
This way you can formalize the sentence "if the cat is on the mat then that cat exists" as
$$
Q(a) \to \exists x (x = a)
$$
where $a$ is that cat, and $Q(y)$ means that "$y$ is on the mat".
Comment. Note that in the standard semantics for first-order logic, it is assumed of every constant symbol is interpreted by on object in the domain of the discourse, which means that$-$in first-order logic with identity$-$the existence predicate $\exists x (x = c)$ is valid for every constant symbol $c$ (see here for a formal proof).

You can refine your formalization: "the mat" in the sentence refers to a specific object in the domain of discourse, so it can be represented by a constant symbol $m$. Therefore, the formalization of the sentence becomes
$$
Q(a,m) \to \exists x (x = a)
$$
where $Q(x,y)$ means that "$x$ is on $y$".

Note that you can also read the sentence as referring to a generic cat, and so its formalization becomes:
$$
\forall y (C(y) \to (Q(y) \to \exists x (x = y)))
$$
where $C(y)$ means that "$y$ is a cat".
A: 
$$Q(a)\rightarrow\exists x$$
or
$$Q(a)\rightarrow\exists a$$
I'm trying to make an argument of the form "if the cat is on the mat then that cat exists"

There are two possibilities:

*

*$a$ is an actual constant (being the cat), in which case the cat is already pre-supposed to exist even before you formulate that malformed sentence. As such, $\text“∃a\text”$ is as meaningful (and useful) as $\text“∃7\text”$ is to mean that the number $7$ exists.

*$a$ is an arbitrary constant or free variable, in which case $Q(a)$ means “[something from the discourse domain, perhaps a cat, but not the cat] is on the mat”. As such, $\text“∃a\text”$ is simply syntactically erroneous and reads as the trailing sentence “for some object in the univese,...”.


Is it not possible then to use symbolic logic to say "if the cat is on the mat then that cat exists"?

Yes, this validity (i.e., first-order tautology, i.e., a logic formula that is true regardless of interpretation) is formalised as $$Q(a)\to\exists x\,(x=a),$$ where the constant $a$ is the cat being remarked about.
A: Theorem
Suppose we have predicates $C$ and $M$ that we interpret as follows:
$~~~~~~C(x)$ = $x$ is a cat
$~~~~~~M(x)$ = $x$ is on the mat
Then we have:
$~~~~~~\forall a:[C(a) \land M(a) \implies \exists b: C (b)]$
Proof

*

*$C(x)\land M(x)~~~~$ (Premise)

*$C(x)~~~~$ (1)

*$\exists b: C(b)~~~~$ (2)

*$\forall a:[C(a)\land M(a) \implies \exists b: C(b)]~~~~$ (Conclusion 1, 3)

