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My professor answered my question on a class discussion board for Discrete Math, but it seems that his answer is just not correct. Could another person who is familiar with this topic weigh in?

For context, the function $C(x)$ means that $x$ is a comedian, and $F(x)$ means that $x$ is funny.

My Question:

Example c was written: $\boldsymbol{∃x(C(x) → F(x))}$

The answer to that example was given as "Someone is a comedian and that means they are funny"

However I would have thought that should translate to: "There is someone who, if they were a comedian, would be funny"

His reply

Those both mean the same thing.

$∃x(P(x) → Q(x))$ never really makes sense....

image of my discussion-board post

ryang
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Edward45
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  • Those two statements mean the same thing to me as well. The "and that means" phrase is meant to convey implication. If the answer were "someone is a comedian and they are funny", I would interpret that as a conjunction. But the phrasing is "and that means", which is different. – John Douma Sep 13 '24 at 00:22
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    @JohnDouma I think the OP is concerned about $x$ not having property $C(x)$ in which case a false statement still implies any other statement. – Severin Schraven Sep 13 '24 at 00:26
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    I agree with you. There is nothing in the statement that tells you that there are any comedians in the world. In fact, the statement is definitely true if there are no comedians in the world, or even if there is a single non-comedian in the world. – lulu Sep 13 '24 at 00:27
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    I agree. To me $\exists x(C(x)\implies F(x))$ is true if there is a single non-comedian in the universe. – Greg Martin Sep 13 '24 at 00:27
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    After all, it is equivalent to $\exists x~(\neg C(x)\lor F(x))$ --- There is someone who is either not a comedian or is funny. – Graham Kemp Sep 13 '24 at 01:32
  • @SeverinSchraven Yes, that makes sense. All Edward has to do, is point out that case to his instructor. That is a pretty clear oversight. – John Douma Sep 13 '24 at 01:51
  • Fun fact: Suppose there exists someone who is NOT a comedian. Then if there exists someone who if they are a comedian then ANY proposition whatsoever must be true! $~~\exists x (\neg C(x)) \implies \exists x (C(x) \implies P)$ for any proposition $P$. – Dan Christensen Sep 13 '24 at 05:10
  • ...If you want to make the domain of discussion explicit where H(x) means x is a human, we have: $~\exists x(H(x) \land \neg C(x))\implies \exists x(H(x) \land (C(x) \implies P))$ for any proposition P. – Dan Christensen Sep 13 '24 at 13:48
  • To be a bit pedantic, $C(x)$ and $F(x)$ are specifically predicates, and just any kind of function. – Mahmoud Sep 13 '24 at 17:23
  • My main problem with the first translation: "and that means they are funny" is something that would typically appear in a proof or an argument, not something you would want to use in specifying what the question is that you want a true or false answer to in the first place. – Daniel Schepler Sep 13 '24 at 22:47

3 Answers3

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Example c was written: $ \exists x (C(x) \to F(x))$

The answer to that example was given as "Someone is a comedian and that means they are funny"

That is an incorrect translation. Imagine a universe with one person who is funny and is not a comedian. Then $\exists x (C(x) \to F(x))$ is true. But it's not true that "someone is a comedian and ..." because nobody is a comedian.

Remember that $(A \to B)$ is equivalent to $((\lnot A) \lor B)$. So we can see that $\exists x (C(x) \to F(x))$ is equivalent to $\exists x ((\lnot C(x)) \lor F(x))$. So a correct translation, if not symbol for symbol, is "there is someone who is not a comedian or is funny". By splitting the existential qualifier, this is also equivalent to $( \exists x (\lnot C(x))) \lor \exists x (F(x))$, "there is someone who is not a comedian, or there is someone funny".

It's not clear exactly what "There is someone who, if they were a comedian, would be funny" means in pure logic, since the subjunctive "were" and "would" sort of brings up multiple "universes".

Tzimmo
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aschepler
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  • I would read "There is someone who, if they were a comedian, would be funny" as "There is someone such that if that person is a comedian, they are funny." This would be equivalent to "There is someone who is either not a comedian or is funny." The notation would be an unusual way to express this, but it's more consistent with how I'd interpret the same statement with a $\forall$ quantifier instead of an $\exists$. – MartianInvader Sep 13 '24 at 20:51
  • @MartianInvader: If you want to properly capture the hypothetical, you need to use modal (alethic) logic, which is well beyond the level of the exercise given. For example, $\exists x \square (C(x) \implies F(x) )$. I don't believe you can eliminate the necessity operator in that expression. – Kevin Sep 14 '24 at 23:45
  • @MartianInvader: Actually, the subjunctive mode "if they were a comedian" exactly captures the logic of this example. And I actually know plenty of such people. – Lee Mosher Sep 15 '24 at 14:23
  • @LeeMosher I agree! I was responding to the last sentence of this answer, since it seems its author doesn't like the subjunctive. I was just trying to point out that the logical interpretation is still fairly straightforward. – MartianInvader Sep 16 '24 at 17:17
  • @MartianInvader: Your logical interpretation has a different meaning, though. "If they were a comedian, they would be funny" implies "but they are not a comedian", but the original logical sentence does not imply this. – psmears Sep 22 '24 at 15:11
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  1. "Someone is a comedian and that means they are funny" actually means $$\exists x\, C(x)\land \forall x\,\big(C(x)\to F(x)\big),$$ which is a logically stronger assertion than $$∃x\,\big(C(x) \to F(x)\big).\tag1$$ For the universe $\{$funny comedian, unfunny comedian$\},$ only the second statement is true.

    Sentence $(1)$ should be translated simply as "There is someone who if they are a comedian then they are funny" rather than "There is someone who, if they were a comedian, would be funny," since we are referring to a specific universe and context.

  2. In the universe $\{$fat dog$\},$ the statements $$∃x\,\big(\operatorname{Cat}(x) → \operatorname{Fat}(x)\big)\\\text{Something is such that }\textit{if }\text{ it's a cat it's fat}\tag1$$ and $$∃x\,\big(\operatorname{Cat}(x) ∧ \operatorname{Fat}(x)\big)\\\text{Some fat cat exists}$$ have opposite truth values.

    $∃x\,\big(C(x) \land F(x)\big)$ is a logically stronger sentence than $∃x\,\big(C(x) → F(x)\big).$

  3. Sentence $(1)$ is actually logically equivalent to $$∃x\,F(x) \;\lor\; \lnot ∀x\,C(x)\\\text{Something fat exists or not everything is a cat}.\tag{1e}$$

P.S. Assertion $P$ being logically stronger than assertion $Q$ means that $P$ logically entails, but isn't logically equivalent to, $Q.$

ryang
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$∃x(P(x) → Q(x))$ never really makes sense.

If $~\exists x (\neg P(x))~$ (usually a reasonable assumption), this will always be true for any predicate $Q$ whatsoever. So, it kind of "makes sense," but I don't know that it will ever be of any use.

In mathematical applications, existential quantifiers are not usually applied to implications. They are usually applied to conjunctions: $\exists x (D(x) \land P(x))$ where $D$ is the domain of quantification, e.g. $~D(x)~\equiv ~ x\in R$.

Dan Christensen
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