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Boris hasn't tried anything better than chocolate.

The above sentence needs to be converted to first-order logic. The given domain is candies, and Boris is a sentient piece of candy.

$B(x,y)$: $x$ is better than $y$

$T(x,y)$: $x$ has tried $y$

$C(x)$: $x$ has chocolate in it

$b$: Boris

I believe the intended meaning is that Boris has not tried any candy that is better than all candies containing chocolate. I have been at this for a few days, and my program says that my attempt $$\forall y\;\forall x\;((C(y)\land B(x,y))\to\lnot T(b,x)) $$ is wrong. I then tried putting $x$ as a negated existential, but I am still struggling to write it correctly.

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    $\begingroup$ Hint: If (Boris has tried y) and (x has chocolate) then y is not better than x. $\endgroup$ May 7, 2023 at 6:02
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    $\begingroup$ The given sentence is unclear since "chocolate" isn't the name of a candy. Does "better than chocolate" mean "better than every candy with chocolate in it" or just "better than at least one candy with chocolate in it"? $\endgroup$
    – Karl
    May 7, 2023 at 6:12
  • $\begingroup$ An accepted answer was @x(@y(Cy > B(x,y)) > -T(b,x)) $\endgroup$
    – MrExp
    May 10, 2023 at 20:39
  • $\begingroup$ @MrExp "An accepted answer was $\forall x\;(\forall y\;(Cy \to Bxy) \to\lnot Tbx).$" $\quad$ In this case, please take a look at Note #3 in my answer below. $\endgroup$
    – ryang
    May 11, 2023 at 16:13
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    $\begingroup$ Ah, I started writing my comment from the 19th in the morning, before you added that formula into your response, and only sent it at night not bothering to check for an edit. Sorry for the confusion, and also touché on your 3-element universe, I didn't think of that possibility. $\endgroup$
    – Nico
    May 23, 2023 at 3:36

2 Answers 2

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The given domain is candies, and Boris is a sentient piece of candy.

Boris hasn't tried anything better than chocolate.

For simplicity, suppose that $$\exists z\;(Cz\land Tbz).$$

The most likely intended meaning:

  • For each candy that Boris has tried, Boris has tried some chocolate that is at least as good as the candy. $$\color{brown}{\forall x\;\big(Tbx\to \exists y\;(Cy\land Tby\land \lnot Bxy)\big)}\tag1$$

Also plausible:

  • For each pair of chocolate and non-chocolate candies that Boris has tried, the chocolate is at least as good as the non-chocolate. $$\color{brown}{\forall y\;\forall x\;(Cy\land Tby\land\lnot Cx\land Tbx\to \lnot Bxy)}\tag2$$

There are two possible literal translations (but problematic: refer to note #3 below):

  • It's not that for some candy that Boris has tried, it is better than every chocolate. $$\lnot\exists x\;\big(Tbx\land\forall y\;(Cy\to Bxy)\big)\tag3$$ For each candy that Boris has tried, some chocolate is at least as good as the candy. $$\color{brown}{\forall x\;\big(Tbx\to \exists y\;(Cy\land \lnot Bxy)\big)}\tag3$$
  • It's not that for some candy that Boris has tried, it is better than some chocolate. $$\lnot\exists x\;\big(Tbx\land\exists y\;(Cy\land Bxy)\big)\tag4$$ For each pair of candy that Boris has tried and chocolate, the chocolate is at least as good as the candy. $$\color{brown}{\forall y\;\forall x\;(Cy\land Tbx\to\lnot Bxy)}\tag4$$

    $$\forall y\;\forall x\;((C(y)\land B(x,y))\to\lnot T(b,x)) \tag4$$


Notes:

  1. The "some chocolate" in assertions $(1)$ and $(3)$ refer to the best chocolate that Boris has tried, and the best chocolate, respectively.

  2. Assertions $(1)$ and $(3),$ but not $(2)$ and $(4),$ allow Boris to have tried a pair of chocolate $p$ and non-chocolate $q$ such that $p$ is worse than $q$.

  3. Assertion $(3)$ allows the best candy that Boris has tried to be non-chocolate while simultaneously having tried chocolate! In this case, none of the other assertions are true.

    Assertion $(4)$ (the OP's suggestion) is so restrictive that it forbids Boris from having tried any chocolate that is better than the worst chocolate! The other assertions allow this.

  4. Under the axiom $$∀p\;¬Bpp \;\land\; ∃z\;(Cz∧Tbz),$$ assertion $(4)$ is logically stronger than the other assertions, while assertion $(2)$ is logically stronger than assertion $(1).$


Addendum corresponding to the OP's new comment

An accepted answer was $∀x\:(∀y\:(Cy→Bxy)→¬Tbx).$"

This is logically equivalent to assertion $(3)$.

Elaborating on Note #3: a surprising issue with assertion $(3)$ is that it is true in a universe with three candies $\text{Choc}_1,\text{NonChoc}_2,\text{Choc}_3,$ where the subscript indicates ranking, only the latter two of which Boris has tried.

Assertion $(3)$ is a literal translation but does not reflect how humans contextually read the given English sentence (due to its inherent ambiguity). (I wonder whether interpreting "Boris hasn't tried anything better than chocolate" as $(1)$ versus $(3)$ is one difference between a smart, well-trained chatbot and an early-generation chatbot.)

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  • $\begingroup$ Can’t Boris like some chocolate more than other chocolate? $\endgroup$
    – PW_246
    May 8, 2023 at 21:56
  • $\begingroup$ @RW_123 Only in the first three assertions. $\endgroup$
    – ryang
    May 11, 2023 at 8:31
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I’d say

$\forall x \forall y [(Tbx \land Tby \land Bxy) \to (Cy \to Cx)]$

That is, “for whatever Boris has tasted, if one thing is better than another, then the one that is worse is chocolate only if the one that is better is chocolate,” or “there is nothing that Boris has tasted that is better than something else he’s tasted such that the better one isn’t chocolate, while the worse one is chocolate.”

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    $\begingroup$ This is equivalent to assertion $(2)$ in my answer. -) $\endgroup$
    – ryang
    May 10, 2023 at 16:38

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