Help With Predicate Calculus Translation

Question

I am trying to learn QL on my own and I wish I could get some help translating the two sentences below.

I have given two sets of answers: Ans A is the textbook one and Ans B is my own attempt. Please tell me whether Ans B is acceptable, and if it's not, what mistake I've made. Thanks!

universe of discourse: candies
$Cx:$ $x$ has chocolate in it.
$Bxy:$ $x$ is better than $y.$

1. The very best candy is chocolate.
2. Any candy with chocolate is better than any candy without it.

Ans A:

1. $∃x[Cx\&¬∃yByx]$
2. $∀x[Cx → ∀y(¬Cy→Bxy)]$

Ans B:

1. $∀x[Cx→¬∃yByx]$
2. $∀x[Cx → ¬∃y(¬Cy\&Byx)]$

Answer 1

universe of discourse: candies
Cx: x has chocolate in it.
Bxy: x is better than y.

1. The very best candy is chocolate.

   textbook: $∃x[Cx \& ¬∃yByx]$
   mine: $∀x[Cx→¬∃yByx]$

Either candies don't exist, or chocolate candy must exist and non-chocolate candy is not better (note that "non-chocolate candy is worse" may be inaccurate). In other words, $$\forall \top \lor \exists x (Cx\land\forall y\lnot Byx);$$ this corresponds to the textbook's answer.

Your answer isn't equivalent to the textbook's, since they have opposite truth values

• in a universe containing only gummy candy, and
• in a universe where pure chocolate candy is better than chocolate-caramel candy.

2. Any candy with chocolate is better than any candy without it.

   textbook: ∀x[Cx → ∀y(¬Cy→Bxy)]
   mine: ∀x[Cx → ¬∃y(¬Cy&Byx)]

Again, the two answers aren't equivalent, since they have opposite truth values

• in a universe where chocolate and gummy candy are equally good.