3
$\begingroup$

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?).

$\endgroup$
7
  • 3
    $\begingroup$ Provided the mat exists. $\endgroup$
    – paw88789
    Jun 30 at 13:14
  • $\begingroup$ $\exists a$ doesn't make sense on its own as a statement that can be true or false. You can only say $\exists a \, P(a)$ where $P(a)$ is some predicate. $\endgroup$ Jun 30 at 13:18
  • $\begingroup$ @paw88789 If the mat didn't exist then "the cat is on the mat" would be false? So "the cat is on the mat" being true entails that both that cat and that mat exist? $\endgroup$ Jun 30 at 13:18
  • 1
    $\begingroup$ @JairTaylor is right. You want something like $Q(a)\to\exists x(x=x)$, ot avoiding the quantifier $Q(a)\to a=a$, which at least mentions the same symbol. (Or, if we also make the mat explicit, $Q(a,\,b)\to a=a\land b=b$ works.) $\endgroup$
    – J.G.
    Jun 30 at 13:19
  • 2
    $\begingroup$ @J.G. did you mean $Q(a)\rightarrow\exists x(x=a)$? Is that something to do with free logic? $\endgroup$ Jun 30 at 13:25

3 Answers 3

2
$\begingroup$

In first-order logic (with identity for point 2):

  1. 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.

  2. 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".

$\endgroup$
4
  • $\begingroup$ Thanks, that looks to be exactly what I was aiming for. $\endgroup$ Jun 30 at 14:30
  • $\begingroup$ @MichaelRushton - You're welcome! $\endgroup$ Jun 30 at 14:32
  • $\begingroup$ Just one other quick question if you don't mind, would this be valid? $¬𝑄(𝑎)→∃𝑥(𝑥=𝑎)$? It's unclear to me if $¬𝑄(𝑎)$ means that there is a cat and it's not on the mat or if it can also mean that there isn't a cat. $\endgroup$ Jun 30 at 14:42
  • $\begingroup$ @MichaelRushton - In the formula $\lnot Q(a) \to \exists x (x = a)$, the negation refers to $Q(a)$ (if you want the negation to refer to the whole formula, you should use parentheses). Therefore, the formula means that "if the cat is not on the mat then that cat exists", and it turns out to be valid as well. Indeed, as I wrote in my answer, the formula $\exists x (x = a)$ is valid, and so every formula of the form $A \to \exists x (x = a)$ is valid as well. $\endgroup$ Jun 30 at 14:51
1
$\begingroup$

$$Q(a)\rightarrow\exists x$$ 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.

$\endgroup$
1
  • $\begingroup$ Is it not possible then to use symbolic logic to say "if the cat is on the mat then that cat exists"? $\endgroup$ Jun 30 at 13:55
1
$\begingroup$

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

  1. $C(x)\land M(x)~~~~$ (Premise)
  2. $C(x)~~~~$ (1)
  3. $\exists b: C(b)~~~~$ (2)
  4. $\forall a:[C(a)\land M(a) \implies \exists b: C(b)]~~~~$ (Conclusion 1, 3)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.