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Can someone help me understand this logic problem:

What are the truth values of these statements?

a) $\exists !\,x\;P(x) \implies \exists x\;P(x)$

b) $\forall x\;P(x) \implies\exists !x\;P(x)$

c) $\exists!x\;\neg P(x) \implies \neg\forall x\;P(x)$

Correct answer (from the book answers):

a) true

b) false

c) true

The way I am see a) for example is:

  1. "There is a unique x for which if $P(x)$ then there exists another $x$ for which $P(x)$"
  2. $\exists !\,x\;P(x)$ can be broken down to just $P(x)$ since there is a "unique $x$"
  3. $\exists x\;P(x)$ can be broken down to: $P(x_1) \lor P(x_2) \lor ... \lor P(x_n)$
  4. Therefore at step #2, if the unique $x$ is FALSE, then the entire statement is TRUE (according to the truth table for $\implies$)
  5. If #2 is TRUE, then step #3 must have at list 1 TRUE $P(x_i)$ for the overall statement to be true,
  6. Therefore I see it can be both TRUE and FALSE depending on $x$. I am not sure how the book comes to true. I don't fully understand this. Can someone correct me and fill my gap in understanding there. Similar issues in understanding b) and c) as well...

Thank you!

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  • $\begingroup$ ad 1: strike "another" $\endgroup$ – Hagen von Eitzen May 21 '18 at 17:55
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    $\begingroup$ The truth of b) depends on the universe, which may have only one object $\endgroup$ – Hagen von Eitzen May 21 '18 at 17:57
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  1. "There is a unique x for which if P(x) then there exists another x for which P(x)"

No. It does not specify that the second $x$ is an other $x$. Also, you make the $\exists !$ to be the main operator of the statement (i.e. you interpret it as $\exists ! x (P(x) \rightarrow \exists x P(x))$ but the main operator is the conditional. So, it says: "If there is exactly one object with property $P$, then there is at least one object with property $P$

  1. "∃!xP(x)" can be broken down to just "P(x)" since there is a "unique x"

No. $P(x)$ is not a claim; it is a formula with a free variable.

  1. "∃xP(x)" can be broken down to: "P(x1) ⋁ P(x2) ⋁ ... ⋁ P(xn)"

... informally that's ok .. but be careful! ... This is assuming $x_1, ..., x_n$ are all the objects in the domain. And they are all constants referring to those objects.

  1. Therefore at step #2, if the unique x is FALSE, then the entire statement is TRUE (according to the truth table for "->")

you don't phrase this very well ('the unique x is FALSE'... ?! .. objects aren't false ...) ... I assume you mean: "If there is no unique object with property $x$, then the whole conditional is true". And yes, that is so.

  1. If #2 is TRUE, then step #3 must have at list 1 TRUE P(xi) for the overall statement to be true

yes ..

  1. Therefore I see it can be both TRUE and FALSE depending on x.

Huh? So you just reasoned that if there is not a unique object with property $P$, then the statement is TRUE, and if there is such a unique object, then the statement is also TRUE ... so how come you say that it could be FALSE or TRUE?

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  • $\begingroup$ Thank you for pointing out my derailments! After analyzing your comments I feel basic understanding on why a) is T. Also see why b) is false unless domain size of 1, and c) is true. $\endgroup$ – jox May 22 '18 at 5:03
  • $\begingroup$ @jox Great! And yes, those answers are all correct. Glad I could help :) $\endgroup$ – Bram28 May 22 '18 at 11:36
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Your thinking is too convoluted.   Just read the statements.

$\exists !x~P(x)\implies \exists x~P(x)$ : "If $P$ holds for exactly one thing, then $P$ holds for something."

Quantfied variables in different scopes may (or may not) refer to the same entities.   They need not be different entities.   (There is no "another" in the statement.)

An entailment is falsified exactly when the consequent may be false while the antecedant is true.

So may it be false that "$P$ holds for something" if it is true that "$P$ holds for exactly one thing"?


$\forall x~P(x)\implies \exists!x~P(x)$ : "If $P$ holds for everything, then $P$ holds for exactly one thing."


$\exists !x~\neg P(x)\implies \neg\forall x~P(x)$ : "If $P$ doesn't hold for exactly one thing, then $P$ holds for not every thing ."

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