Can we ever use implication with an existential quantifier?

Question

I am doing a class on logic and I was given the following question: Is this predicate a valid formalization of "some dogs are sleepy"?

The statement in question is

$\exists d \in Dogs: is\,a \,dog(d) \implies sleepy(d)$

now I know that the rule of thumb is that we should use the conjunction. My understanding of the question was that $is\,a\,dog(d)$ is a tautology based on the previous predicate bounding $d$ to $Dogs$. This would result in the following predicate:

$\exists d \in Dogs: True \implies sleepy(d)$

then this would be equivalent to:

$\exists d \in Dogs: sleepy(d)$

giving interpreting the resulting predicates gives the meaning: "some dogs are sleepy". Can modus ponens be used in predicate to infer that two predicates are equivalent, thus giving them the same meaning?

Answer 1

No, the correct formalization is:

$\exists d (Dog(d) \land Sleepy(d))$.

You can abbreviate it as :

$(\exists d \in Dogs) (Sleepy(d))$.

Why your proposal does not work ?

Because if we change "universe" and we move to that of $Men$, we have that

$\exists d (Dog(d) \land Sleepy(d))$

is false: no memeber of the universe is a dog and thus $Dog(x) \land Sleepy(x)$ is false for every value, while:

$\exists d (Dog(d) \to Sleepy(d))$

will be true, because $Dog(x) \to Sleepy(x)$ is vacuously true.

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"Two predicate [logic formula]s are equivalent"

when they have always the same truth values (i.e. either both true or both false) where "always" means: in every possible interpretation.

Answer 2

The notation you're using, $$ \exists d \in Dogs: isAdog(d) \implies sleepy(d)$$ can be viewed as "syntactic sugar" for the expression $$ \exists d ((d \in Dogs) \land (isAdog(d) \implies sleepy(d))).\tag1$$

Moreover, the name $isAdog$ seems superfluous, since apparently $isAdog(d) \equiv (d \in Dogs)$, and $(d \in Dogs)$ is just about as easy to write as $isAdog(d).$ So Predicate $1$ is equivalent to $$ \exists d ((d \in Dogs) \land ((d \in Dogs) \implies sleepy(d))). \tag2$$

And then I think you are correct: $(d \in Dogs) \land ((d \in Dogs) \implies sleepy(d))$ can be replaced by $(d \in Dogs) \land sleepy(d)$ (or vice versa), so the Predicate $2$ is completely equivalent to $$ \exists d ((d \in Dogs) \land sleepy(d)). \tag3$$

But why would you translate "some dogs are sleepy" into Predicate $1$ or Predicate $2,$ or even a disguised form of one of these predicates using the $\exists d\in Dogs$ notation, when you can instead write Predicate $3$?

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If we really want to express "some dogs are sleepy" using an (slightly) excessively complex predicate with an implication, we could write

$$ \exists d (\lnot((d \in Dogs) \implies \lnot sleepy(d))),$$

using the fact that $A \land B$ is equivalent to $\lnot(A\implies\lnot B).$

Answer 3

Is this predicate a valid formalization of "some dogs are sleepy"?

The statement in question is

$\exists d \in Dogs: isAdog(d) \implies sleepy(d)$

Assuming you meant $d\in Dogs \equiv isAdog(d)$, equivalent to your original statement would be:

$\exists d:[isAdog(d) \land [isAdog(d) \implies sleepy(d)]]$

Equivalent, too, would be:

$\exists d: [isAdog(d) \land sleepy(d)]$

(Proof left as an exercise.)

So, there is nothing wrong with your original statement, but I think you will agree this is a much more straightforward formalizing "some dogs are sleepy."

...now I know that the rule of thumb is that we should use the conjunction.

There is a good reason for that, especially in set theory (see my comments), but you actually have an existential quantifier on a conjunction.

My understanding of the question was that $isAdog(d)$ is a tautology based on the previous predicate bounding $d$ to $Dogs$. This would result in the following predicate:

$\exists d \in Dogs: True \implies sleepy(d)$

I am not familiar with this usage of "True."

...then this would be equivalent to:

$\exists d \in Dogs: sleepy(d)$

This, too, is equivalent to the original statement.