I was reading chapter-1 The Foundations: Logic and Proofs from this book.

The chapter gives example of translating English sentence : "There is a woman who has taken a flight on every airline in the world." as follows:

  • Introducing variables : w for women, f for flight, a for airline
  • Let P(w,f) : “w has taken f”
  • Let Q(f,a) : “f is a flight on a.”
  • Translation : ∃w∀a∃f (P(w,f ) ∧ Q(f,a))

I did understood above. Next it gives example of translating negation of above sentence : "There does not exist a woman who has taken a flight on every airline in the world.", which it solved as follows:

¬∃w∀a∃f (P(w,f ) ∧ Q(f,a)) ≡ ∀w¬∀a∃f (P(w, f ) ∧ Q(f, a))

                            ≡ ∀w∃a¬∃f (P(w, f ) ∧ Q(f, a))

                            ≡ ∀w∃a∀f¬(P (w, f ) ∧ Q(f, a))

                            ≡ ∀w∃a∀f (¬P(w, f )∨¬Q(f, a))

I thought their can be straight approach for this translation instead of going through negation. Anyways though my problem is I am unable to interpret the final translation ∀w∃a∀f (¬P(w, f )∨¬Q(f, a)) in plain English.

"Every woman has either not taken all flights or out of all the flight she has taken are not there on some airline." Is this correct? But if yess, it still does not make me much sense. Anyone?

  • $\begingroup$ I dont know if this is somewhat not very precise question and I may be at risk of getting banned if this question is closed for not being precise cause cs.stackexchange is already warning me. But logic always proves fuzzy to me and making a confusing logic more clear is itself an effort. So please help. I wish there could have been some place on stackexchange where I can ask my initial questions and promote them to actual questions if they are valid. Chat rooms are there but cs chatroom is not active. $\endgroup$
    – Maha
    Oct 11 '14 at 10:34
  • $\begingroup$ There's a reason we use formal logic rather than English for stating complex mathematical propositions: English is imprecise and not a great tool for that purpose. $\endgroup$
    – D.W.
    Oct 11 '14 at 18:57

The negation is that every woman has an airline that she has never flown with. This is equivalent to saying that, for every woman there's an airline with the following property:

  • for every flight in the whole world, either the woman was not on that flight (regardless of what airline operated it) or the flight was not operated by that airline (regardless of whether the woman was on it or not).

That is, there was no flight by the airline in question that the woman was on.

This is not quite the same as what you have.

  • $\begingroup$ wow...first at least that sounds a bit logical and straight translation from logic to English, am still not able to fully comprehend it $\endgroup$
    – Maha
    Oct 11 '14 at 13:08
  • $\begingroup$ ok I think I got it little bit more, is the above interpretation correct? $\endgroup$
    – Maha
    Oct 11 '14 at 13:19
  • $\begingroup$ I accepted your edit by mistake. It was not correct. "There is an airline to which the flight does not belong" is true for every flight. Every flight is not a British Airways flight, except for BA's flights, which are not Delta Air Lines flights. $\endgroup$ Oct 11 '14 at 15:22

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