Logics with Verbs, Relative Clauses, and Comparative Adjectives

Transcription

1 1/48 Logics with Verbs, Relative Clauses, and Comparative Adjectives Larry Moss, Indiana University EASLLC 2014

2 2/48 Where we are FOL Turing FO 2 + trans RC (tr,opp) Aristotle FO 2 S S R R RC (tr) RC RC(tr) RC RC(tr, opp) adds full N-negation R + (transitive) comparative adjs R + relative clauses done relational syllogistic all these are today done

3 3/48 Preliminary work, done as a set of group exercises I am going to introduce this topic in a slow way by cutting things down to a very small language and working up. (This is my favorite thing to do for this subject!) Working with the small language enables us to do a lot of the proofs, and to get our hands dirty. After we do this, I will go more quickly and omit a huge number of details.

4 4/48 Does the conclusion follow? All boys are males All women see all males All who see all boys see all males The definition of follow here is in terms of models. For this fragment, a model M should be a set M, the universe subsets of M for the nouns: [[boys]], [[males]], and [[females]] a subset of M M for the verb: [[see]] In other words, it is a model like we had before, but with a set of pairs of individuals to interpret the verb.

5 5/48 Syntax We make set terms in the following way: see is a set term. If x is a set term, so is (see all x). We can read this in various ways to make it sound more like English. Sometimes it would be as see all who are x. So we get set terms like see all boys see all (see all women) see all (see all (see all women))

6 5/48 Syntax We make set terms in the following way: see is a set term. If x is a set term, so is (see all x). We can read this in various ways to make it sound more like English. Sometimes it would be as see all who are x. So we get set terms like see all boys see all (see all women) see all (see all (see all women)) Later on, we ll call set terms set terms.

7 6/48 Syntax and Semantics For our sentences, we take all expressions where x and y are set terms. All x y In the semantics, we need an inductive definition to interpret the set terms. [[see all x]] = {m M : }

8 6/48 For our sentences, we take all expressions where x and y are set terms. All x y Syntax and Semantics In the semantics, we need an inductive definition to interpret the set terms. [[see all x]] = {m M : } [[see all x ]] = {m M : for all n [[x]], (m,n) [[see]]}

9 7/48 Example of a model [[males]] = {a,d,f } [[females]] = {b,c,e} [[boys]] = {a,d} d c b a e f Technically, M = {a,...,f }, [[see]] = {(a,b),(a,c),(a,e),(c,d),(b,a),(b,b), (b,f ),(c,f ),(d,b),(d,d),(d,e),(d,f )}

10 Example of a model [[males]] = {a,d,f } [[females]] = {b,c,e} [[boys]] = {a,d} d c b a e f What is [[see all males]]? What is [[see all females]]? What is [[see all (see all females)]]? What is [[see all (see all (see all females))]]? 7/48

11 8/48 Logic For sure we need the logic of All from earlier: All x are x All x are y All y are z All x are z Barbara But note that we are using this with x, y, and z as set terms. But in fact we are missing at least one rule!

12 9/48 Logic All x are x axiom All y x All (see all x)(see all y) skunk All x are y All y are z All x are y Barbara

13 9/48 Logic All x are x axiom All y x All (see all x)(see all y) skunk All x are y All y are z All x are y Barbara Example: All x (see all y),all z y All x (see all z) All z y All x (see all y) All (see all y)(see all z) All x (see all z)

14 10/48 Does the conclusion follow? All boys are males All who see all women also see all who see all boys All who see all see all boys they are all women All women see all males All who see all boys see all males In our language, this is All boys males All (see all women) (see all (see all boys)) All (see all (see all boys)) women All women (see all males) All (see all boys) (see all males)

15 11/48 The canonical model Γ of a set of assertions in this logic We are given a set Γ in this language. We aim to build a model M of Γ with the property that if M = ϕ, then Γ ϕ. This would prove the completeness theorem for our logic, because if Γ ϕ, then M = ϕ. (In other words, every sentence which is true in all models of Γ (hence in M) would be provable from Γ.)

16 12/48 The canonical model Γ of a set of assertions in this logic Let M be the set of all set terms. Let [[females]] = {x : Γ All x females} [[males]] = {x : Γ All x males} [[boys]] = {x : Γ All x boys} The hardest point is to decide on the semantics of see

17 12/48 The canonical model Γ of a set of assertions in this logic Let M be the set of all set terms. Let [[females]] = {x : Γ All x females} [[males]] = {x : Γ All x males} [[boys]] = {x : Γ All x boys} The hardest point is to decide on the semantics of see [[see]] = {(x,y) : Γ All x see all y}

18 Does the conclusion follow? All p are q All who see all who see all p (see all q) The way we ll do this is to draw the canonical model of Γ, where Γ = {All p are q}. To make the notation simpler, let us write p 0 for p, and p n+1 for see all p n ; we adopt similar notation for q. Then the canonical model of Γ is shown below: p 0 p 1 p 2 p 3 p 4 q 0 q 1 q 2 q 3 q 4 Notice that the notation for subset arrows is different than the arrows for see. Notice also that we have omitted the reflexive subset arrows on all of the nodes, and also that the arrows which are shown alternate directions. 13/48

19 14/48 Theory Lemma: [[x]] = {y : Γ All y are x} The proof is by induction on x. Lemma: M = Γ This follows from the last lemma. Lemma: If M = ϕ, then Γ ϕ. This is basically what we saw yesterday for the language of All.

20 15/48 Solving the problem Does this follow? All boys males All (see all women) (see all (see all boys)) All (see all (see all boys)) women All women (see all males) All (see all boys) (see all males) We make the part of the canonical model just using the set terms above, and all the subterms: boys men women see all boys see all men see all women see all see all women

21 16/48 More boys see all boys subset see all see all boys subset subset see all males see all females males subset females

22 17/48 Here is part of the see relation The logic gives us part of the see relation in a trivial way boys see all boys see all see all boys see all males see all females males females

23 18/48 Many verbs What would we do if we had more than one verb? All women are football players All football fans love all football players All football fans like all women Does this work?

24 19/48 Background knowledge In addition to our assumptions Γ, we can adopt a set of background assumptions which need not be expressible in our language. In this case, we might say = {loves likes} Then we require that our models respect the background assumptions: [[loves]] [[likes]].

25 20/48 Background knowledge: a new logical law All x love all y All x like all y What does completeness say? And why does it hold?

26 20/48 Background knowledge: a new logical law All x love all y All x like all y What does completeness say? Theorem: The following are equivalent Every model of Γ which respects the assumptions in satisfies ϕ. There is a proof of ϕ using the assumptions in Γ and the logic previously described, augmented by the inference rules corresponding to the sentences in And why does it hold?

27 Adding transitive verbs all work onrand related systems is joint with Ian Pratt-Hartmann The next language uses see or r as variables for transitive verbs. All p are q Some p are q All p see all q All p see some q Some p see all q Some p see some q All p aren t q No p are q Some p aren t q All p don t see all q No p sees any q All p don t see some q No p sees all q Some p don t see any q Some p don t see some q The interpretation is the natural one, using the subject wide scope readings in the ambiguous cases. This is R. (The first system of its kind was Nishihara, Morita, Iwata 1990.) The language R has complemented atoms p on top of R. 21/48

28 22/48 Example of a proof in the system for R What do you think? Sound or unsound? All X see all Y,All X see some Z,All Z see some Y = All X see some Y

29 22/48 Example of a proof in the system for R What do you think? Sound or unsound? All X see all Y,All X see some Z,All Z see some Y = All X see some Y The conclusion does indeed follow: take cases as to whether or not there are Z. We should have a formal proof.

30 23/48 Example of a proof in this system All X see all Y,All X see some Z,All Z see some Y All X see some Y Some X see no Y Some X are X All X see some Z Some X see some Z Some Z are Z All Z see some Y Some Z see some Y Some Y are Y All X see all Y All X see some Y Some X see no Y Some X aren t X

31 24/48 But now [Some X see no Y ] Some X are X All X see some Z Some X see some Z Some Z are Z All Z see some Y Some Z see some Y Some Y are Y All X see all Y All X see some Y [Some X see no Y ] Some X aren t X All X see some Y RAA This shows that All X see all Y,All X see some Z,All Z see some Y All X see some Y

32 25/48 Towards the syntax for R All p are q (p,q) Some p are q (p,q) All p r all q (p, (q,r)) All p r some q (p, (q,r)) Some p r all q (p, (q,r)) Some p r some q (p, (q,r)) No p are q (p,q) Some p aren t q (p,q) All p don t r all q No p r any q (p, (q,r)) All p don t r some q No p r all q (p, (q,r)) Some p don t r any q (p, (q, r)) Some p don t r some q (p, (q, r))

33 25/48 Towards the syntax for R All p are q (p,q) Some p are q (p,q) All p r all q (p, (q,r)) All p r some q (p, (q,r)) Some p r all q (p, (q,r)) Some p r some q (p, (q,r)) No p are q (p,q) Some p aren t q (p,q) No p r any q (p, (q,r)) No p r all q (p, (q,r)) Some p don t r any q (p, (q, r)) Some p don t r some q (p, (q, r)) set terms c positive p (p, r) (p, r) negative p (p, r) (p, r)

34 26/48 Reading the set terms (p,r) those wh r all p (p,r) those wh r some p (p,r) those wh fail-to-r all p = those wh r no p (p,r) those wh fail-to-r some p = those wh don t r some p

35 27/48 More on the set terms To understand how they work, let us exhibit a rendering of the simplest set terms into more idiomatic English: (boy,see) (girl,admire) (boy,see) (girl,recognize) those who see all boys those who admire some girl(s) those who fail-to-see all boys = those who see no boys = those who don t see any boys those who fail-to-recognize some girl = those who don t recognize some girl or other

36 28/48 Towards the syntax for R All p are q (p,q) Some p are q (p,q) All p r all q (p, (q,r)) All p r some q (p, (q,r)) Some p r all q (p, (q,r)) Some p r some q (p, (q,r)) No p are q (p,q) Some p aren t q (p,q) No p sees any q (p, (q, r)) No p sees all q (p, (q, r)) Some p don t r any q (p, (q, r)) Some p don t r some q (p, (q, r)) simplifies to (p, c) (p, c) set terms c positive p (p, r) (p, r) negative p (p, r) (p, r)

37 29/48 Syntax of R and related languages We start with one collection of unary atoms (for nouns) and another of binary atoms (for transitive verbs). expression variables syntax unary atom p, q binary atom r positive set term c + p (p,r) (p,r) set term c,d p (p,r) (p,r) p (p,r) (p,r) R sentence ϕ (p, c) (p, c) R sentence ϕ (p,c) (p,c) (p,c) (p,c) RC sentence ϕ (d +,c) (d +,c) RC sentence ϕ (d,c) (d,c)

38 30/48 Negations We need one last concept, syntactic negation: expression syntax negation positive set term c p p p p (p,r) (p,r) (p,r) (p,r) (p,r) (p,r) (p,r) (p,r) R sentence ϕ (p, c) (p, c) (p, c) (p, c) Note that p = p, c = c and ϕ = ϕ.

39 example Consider the model M with M = {w,x,y,z}, [[cat]] = {w,x,y}, [[dog]] = {z}, with [[see]] shown below on the left, and [[likes]] on the right: w x y z x,z y w Then [[ (dog,see)]] is the set of entities that see some dog, namely {x,y,z,w}. Similarly, [[ (dog,likes)]] = {w,y}. It follows that Since [[cat]] contains x, we have [[ ( (dog,likes),see)]] = {x}. M = ( ( (dog, likes), see), cat). That is, in our model it is true that everything which sees everything bigger than some dog is a cat. 31/48

40 32/48 Relational syllogistic logic p and q range over unary atoms, c over set terms, and t over binary atoms or their negations. (p, q) (q, c) (p, c) (p, q) (q, c) (p, c) (p, q) (p, c) (q, c) (p, p) (p, c) (p, p) (q, c) (p, c) (p, q) (p, p) (p, c) (p, (q, t)) (q, q) (p, (n, t)) (q, n) (p, (q, t)) (p, (q, t)) (q, n) (p, (n, t)) (p, (q, t)) (q, n) (p, (n, t)) [ϕ]. (p, p) RAA ϕ

41 33/48 Relational syllogistic logic Most are monotonicty principles (p,q ) (p,q ) (p, (q,t)) (p, (q,t)) (p, (q,t)) (p, (q,t)) Plus also (p, p) (p, c) (p, p) (p, p) (p, c) (p, (q, t)) (q, q) (q, c) (p, c) (p, q) ( ) (p, (n, t)) (q, n) (p, (q, t)) Of these, ( ) is the most interesting.

42 34/48 Example of a proof in the system for R Every porter recognizes every porter No quarterback recognizes any quarterback No porter is a quarterback The derivation in the logical system for R is (q, (q,r)) (p, (p,r)) [ (p,q)] (p, (q,r)) [ (p,q)] (q, (q,r)) ( ) (q, q) (p,q) RAA

43 35/48 Results on R and R Theorem The system that we have is complete. There are no purely syllogistic logical systems complete for R, that is no systems without RAA.

44 35/48 Results on R and R Theorem The system that we have is complete. There are no purely syllogistic logical systems complete for R, that is no systems without RAA. Theorem There are no finite, complete syllogistic logical systems for R, even ones which allow RAA.

45 36/48 RCA Now let s add comparative adjectives, interpreted as transitive relations.

46 37/48 Example of what we want to formalize in RCA Example Every hyena is taller than some jackal Everything taller than some jackal is not heavier than any warthog Everything which is taller than some hyena is not heavier than any warthog

47 38/48 Syntax expression variables syntax unary atom p, q adjective atom a tv atom v binary atom r v a positive set term c +,d + p (p,r) (p,r) set term c,d c + p (p,r) (p,r) sentence ϕ (d +,c) (d +,c)

48 39/48 A few more examples (boy,see) (girl,taller) (boy,see) (girl,see) those who see all boys those who are taller than some girl(s) those who fail-to-see all boys = those who see no boys = those who don t see any boys those who fail-to-see some girl = those who don t see some girl

49 40/48 The set terms in this language are a recursive construct We may embed set terms. So we have set terms like ( (cat, sees), taller) which may be taken to denote the individuals who are taller than someone who sees no cat. We should note that the relative clauses which can be obtained in this way are all missing the subject, never missing the object. The language is too poor to express predicates like λx.all boys see x.

50 41/48 Semantics A model (for this language L) is a pair M = (M,[[ ]]), where M is a non-empty set, [[p]] M for all p P, [[r]] M 2 for all binary atoms r R. The only requirement is that for all adjectives a, [[a]] should be a transitive relation: if a(x,y) and a(y,z), then a(x,z).

51 42/48 Semantics Given a model M, we extend the interpretation function [[ ]] to the rest of the language by setting [[p]] = M \ [[p]] [[r]] = M 2 \ [[r]] [[ (l,t)]] = {x M : for some y such that [[l]](y), [[t]](x,y)} [[ (l,t)]] = {x M : for all y such that [[l]](y), [[t]](x,y)} We define the truth relation = between models and sentences by: M = (c,d) iff [[c]] [[d]] M = (c,d) iff [[c]] [[d]] If Γ is a set of formulas, we write M = Γ if for all ϕ Γ, M = ϕ.

52 Logic (c,c) (T) (c,d) (c,c) (I) (c,b) (b,c) (C) (b,c) (c,d) (B) (b, d) (b, c) (c, d) (b, d) (D1) (b, c) (b, d) (c, d) (D2) (p, q) ( (q,r), (p,r)) (J) (p, q) ( (p,r), (q,r)) (L) (p, p) (c, (p,r)) (Z) (p, (q, a)) ( (p,a), (q,a)) (tr1) (p, (q, a)) ( (p,a), (q,a)) (tr3) (p,q) ( (p,r), (q,r)) (K) (q, (p,r)) (p, p) (II) (p, p) ( (p,r), (p,r)) (W) (p, (q,a)) ( (p,a), (q,a)) (tr2) (p, (q,a)) ( (p,a), (q,a)) (tr4) [ϕ]. ϕ RAA The proof system for logic of RCA. In it, p and q range over unary atoms, b and c over set terms, d over positive set terms, r over binary atoms, and a over adjective atoms. 43/48

53 44/48 An invalid inference The following inference is invalid: Every giraffe sees every gnu Some gnu sees every lion Some lion sees some zebra Every giraffe sees some zebra To see this, consider the model shown below giraffe gnu lion zebra The interpretations of the unary atoms are obvious, and the interpretation of the verb is the relation indicated by the arrow.

54 45/48 Reading the rules (J) If all watches are gold items, then everyone who owns all gold items owns all watches. (K) If all watches are gold items, then everyone who owns some watch owns some gold item. (L) If some watches are gold items, then everyone who owns all watches owns some gold item. (II) If someone owns a watch, then there is a watch. (tr1) If all watches are bigger than some pencil, then everthing bigger than some watch is bigger than some pencil. (tr2) If all watches are bigger than all pencils, then everthing bigger than some watch is bigger than all pencils. (tr3) If some watch is bigger than all pencils, then everthing bigger than all watches is bigger than all pencils. (tr4) If some watch is bigger than some pencil, then everthing bigger than all watches is bigger than some pencil.

55 46/48 Return of the skunks Here is a derivation for an example from early on in this course: (skunk, mammal) ( (mammal,respect), (skunk,respect)) (J) ( ( (skunk,respect),fear), ( (mammal,respect),fear)) (J) Note that inference using (J) is antitone each time: skunk and mammal have switched positions.

56 47/48 Return of the jackals and hyenas Every hyena is taller than some jackal Everything taller than some jackal is not heavier than any warthog Everything which is taller than some hyena is not heavier than any warthog (hyena, (jackal, taller)) ( (hyena,taller), (jkl,taller)) (tr1) ( (jkl,taller), (warthog,heavier)) ( (hyena, taller), (warthog, heavier)))

57 48/48 Zero Here is a proof of the (Zero) rule of S : (p,p) (p,q): [ (p,q)] 1 (p, p) (I) (p, p) (p, p) (p,q) (RAA)1 (D1)