I have to translate the following English sentences into First-Order Logic without using quantifiers:

  1. Everyone on flight 815 has a story.
  2. No one knows what is inside the hatch.
  3. Someone on the island isn't on the flight manifest.

I have tried it, but can't translate without using ∀ and ∃:

  1. $\forall x, \; \text{flight815}(x) \rightarrow \text{story}(x)$
  2. $\forall x \neg(\text{knows}(x) \rightarrow \text{inside hatch}(x))$ // not sure about this OR $\neg \exists x, \; \text{knows}(x, \text{inside hatch})$
  3. $\exists x, \; \text{island}(x) \land \neg(\text{flight manifest}(x))$

Is it possible to do it? If not, why?

Refer to chapter 8 of Artificial Intelligence: A Modern Approach (3rd edition). Stuart Russell and Peter Norvig, Prentice Hall (2010)

  • $\begingroup$ It's not possible to do it since the sentences which you gave are complex sentences. Words like "Everyone", "All", "Somebody", "At least", etc. are supported only by FOL and not by simple propositional logic! $\endgroup$
    – kiner_shah
    Commented Feb 24, 2017 at 4:55
  • $\begingroup$ @kiner_shah Thank so so much. Do you think I have translated it correctly to FOL. I am confused about the second sentence. $\endgroup$
    – New_Coder
    Commented Feb 25, 2017 at 13:31
  • $\begingroup$ @New_Coder, I am not sure about the second FOL sentence. Try forming the sentence: "Everybody knows what's inside the hatch" (It could be something like "for all x, if knows(x) then there exists y such that y is inside the hatch") and then figuring out how to modify the FOL to fit your second sentence. $\endgroup$
    – kiner_shah
    Commented Feb 26, 2017 at 5:29
  • $\begingroup$ @kiner_shah Is this looks better for 2nd sentence: ∀x ⌐(knows(x) → ∃y inside hatch(y)) I have also changed the 1st sentence: ∀x fight815(x) → ∃y story(y) What do you think now? There's an example I found at: uobabylon.edu.iq/eprints/publication_5_29514_1380.pdf No student loves Bill: ¬ ∃ x ( Student(x) ∧ Loves(x, Bill) ) $\endgroup$
    – New_Coder
    Commented Feb 26, 2017 at 14:24
  • $\begingroup$ 1. ∀x, onFlight815(x) -> hasStory(x) 2. ~(∃x,y knows(x,y) ^ insideHatch(y)) What about this? $\endgroup$
    – kiner_shah
    Commented Feb 27, 2017 at 8:58


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