# Help With Predicate Calculus Translation

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. $$∀x[Cx → ¬∃y(¬Cy\&Byx)]$$
• B1 is wrong. It must be: $¬∀x[Cx→∃yByx]$ Sep 20 at 10:36
• B2 must be: $∀x[Cx → ¬∃y(¬Cy \& ¬Byx)]$ Sep 20 at 10:44
• @MauroALLEGRANZA would you be able to say $( \forall x)( \forall y)[(Cx & Cy) \rightarrow Bxy]$ for B2? Sep 23 at 16:51
• Welcome to the site! If the problem has been resolved, consider accepting and/or upvoting answers: it scores points, signals resolution, and prevents bumping. Oct 7 at 7:28

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.
1. 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.