Is this a correct translation of english to first order logic?

Asked Oct 02 '22 at 19:35

Active Oct 02 '22 at 20:25

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Working through first order logic examples and want to make sure I am translating correctly before moving on to harder resolution of first order logic

Use the predicates HasBugs(x), Comments(x), HighQuality(x), QuickSolve(x), and BestPractice(x), where x is a programmer

My translations are:

"All programmers who do not have bugs in their code and comment their code write high-quality code"
∀x(¬HasBugs(x) ∧ Comments(x) ⇒ HighQuality(x))

"Programmers who use best coding practices comment their code"
∀x(BestPractice(x) ⇒ Comments(x))

"Programmers who write high-quality code solve programming problems quickly."
∀x(HighQuality(x) ⇒ QuickSolve(x))

"Amanda uses best coding practices and does not have bugs in her code."
BestPractice(Amanda) ∧ ¬HasBugs(Amanda)

"Bill does not comment his code."
¬Comments(Bill)

Note: my main question is if I have to include a Programmer(x) predicate at all or if my use of Amanda is correct here? Or is this implied since x is a programmer?

edited Oct 02 '22 at 20:25

asked Oct 02 '22 at 19:35

RTT

1

Looks OK to me. – Ethan Bolker Oct 02 '22 at 19:38
I agree correct. – coffeemath Oct 02 '22 at 19:40
I think it fails if $x$ is not a programmer. That's why I think it is best condition to include Programmer(x) predicate as a condition. That's my personal prefference when translating this type of sentences. – MathStackExchange Oct 02 '22 at 19:45
@MathStackExchange So for #3, would I write it as ∀x(Programmer(x) ⇒ (HighQuality(x) ⇒ QuickSolve(x))). I have a strong feeling that is wrong but im not sure how to include it correctly – RTT Oct 02 '22 at 19:46
1

@RTT Indeed, $\forall$ quantor matches with $\to$ operator in conditional contexts. – MathStackExchange Oct 02 '22 at 19:48

Is this a correct translation of english to first order logic?

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