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