I have a question if First Order Predicate Logic always has to include quantifiers? E.g. in the sentence, A black dog bit a small child, would it be:

Bit(dog, child)


∀𝑥: Bit (dog, child) ?

Therefore, in the sentence All the children who live in Otley know each other, would it be:

∃𝑥: Know(x, y)


∃𝑥: Know (x, child) ?


Fill in the blanks with suitable predicates.

Example 1

There exists x and y such that BlackDog(x) and ___________________ and ________________.

Example 2

For all x and y, if OtleyChild(x) and _________________ then ___________________ and ______________________.

  • $\begingroup$ Okay I'm not sure, I'd say in the first Ex1 it would be 'and Child(y) and Bite(x,y)'? And in Ex2 'and know(x,y) then know(y, x)'? Is it correct? $\endgroup$ – Ala Pa May 7 at 23:20
  • $\begingroup$ Example 1 OK. I hadn't considered that "knowing" doesn't necessarily work both ways in Example 2. Well done! Made edit to Example 2 for extra predicate. You still need to fill in the first one. $\endgroup$ – Dan Christensen May 8 at 1:53
  • $\begingroup$ So bascially then the first example is ∀𝑥∀y: Child(y) ^ BlackDog(x) -> Bite(x,y)? And in the second example after the edit, would it be - "and OtleyChild(y) then know(x,y) and know(y,x)"? So ∃𝑥∃y: OtleyChild(y) -> Know(x,y) and know(y,x)? $\endgroup$ – Ala Pa May 8 at 23:53
  • $\begingroup$ Nope. $\forall 𝑥~\forall y: \text{Child}(y) \land \text{BlackDog}(x) \to \text{Bite}(x,y)$ says "Every black dog bit every child". However, you want to say "A black dog bit a small child." $\endgroup$ – Graham Kemp May 9 at 0:11
  • $\begingroup$ Oh, so it would be just ∃𝑥∃y :Child(𝑦)∧BlackDog(𝑥)→Bite(𝑥,𝑦)? Sorry, didn't notice this mistake. How about the other one? Is it ∃𝑥∃y: OtleyChild(y) -> Know(x,y) ^ know(y,x)? $\endgroup$ – Ala Pa May 9 at 18:49

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