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I need some help translating the following English sentences to predicate logic. I want to make sure I'm doing it correctly.

a) Any pet either loves itself or some other person.

What I got: $$\forall x\;\Big(\text{Pet}(x) \Longrightarrow \exists y\big(\text{Loves}(x,x) \lor \text{Loves} (x,y)\big)\Big)$$


b) Dogs will eat anything.

What I got: $$ \forall x\;\forall y\;\big(\text{Dog(x)} \Longrightarrow \text{willEat}(x,y)\big) $$


c) Some sleepy student didn't answer any questions.

What I got: $$ \exists x\;\big((\text{student}(x) \land \text{sleepy}(y)) \Longrightarrow \lnot \text{answeredQuestion}(x)\big) $$

Should this one be the following? $$ \exists x\;\big((\text{student}(x) \land \text{sleepy}(y)) \land \lnot \text{answeredQuestion}(x)\big) $$


d) No dog except Fido barked.

What I got: $$\forall x\;\big(\text{dog}(x) \Longrightarrow (\text{Fido}(x) \Longleftrightarrow \text{barked}(x)\big) $$


Are any of these incorrect or do I seem to be doing it OK?

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  • $\begingroup$ Note that, in this case it would feel more natural to use $\to $ instead of $\implies $. See math.stackexchange.com/a/1616379/53301 for further information. $\endgroup$ – JnxF Jan 27 '16 at 22:57
  • $\begingroup$ (Supplement to @JnxF 's comment) That's "\to". Similarly, the corresponding symbol for equivalence in first order formulas is $\leftrightarrow$, "\leftrightarrow". (What you probably wanted to use was not the two arrows side by side but rather $\Leftrightarrow$, which is "\Leftrightarrow", or $\Longleftrightarrow$, which is "\Longleftrightarrow".) $\endgroup$ – BrianO Jan 27 '16 at 23:00
  • $\begingroup$ Whoops, noted. Thanks! $\endgroup$ – dibdub Jan 27 '16 at 23:14
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a), b) and d) are correct (assuming d) implies that Fido did in fact bark.)

For c), the second form with the conjunction is closer to correct, but, owing to at least a typo [$sleepy(y)$ — what is $y$?], not quite. It should be:

$ \exists x(student(x) \land sleepy(x) \land \forall y\,(isQuestion(y)\to \neg\, answered(x, y)).$

I decomposed $answeredQuestions(x)$ into simpler predicates $isQuestion(y)$ and $answered(x,y)$ in order to reveal more of the structure of the sentence.

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  • $\begingroup$ Awesome, that makes sense! Yeah, the y in part c was just a typo, that was supposed to be x. Thanks for the help! $\endgroup$ – dibdub Jan 27 '16 at 22:57
  • $\begingroup$ No prob, you're welcome. $\endgroup$ – BrianO Jan 27 '16 at 23:04

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