Relational Reasoning in Natural Language

Transcription

1 1/67 Relational Reasoning in Natural Language Larry Moss ESSLLI 10 Course on Logics for Natural Language Inference August, 2010

2 Adding transitive verbs the work on R, R, and other systems is joint with Ian Pratt-Hartmann The main language in this lecture 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.) Another language called R has complemented atoms p on top of R. 2/67

3 3/67 Semantics [[r]] M M Issue: is there one verb, or many? For this fragment, we might as well restrict attention to just one verb. But when we move to the fragment with relative clauses, this will not do.

4 4/67 Are there any interesting inferences? { No x see any y All z see some y } = No x are z All x see all y All p are y Some p are q = All x see some q

5 5/67 Syntax of R 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))

6 5/67 Syntax of R 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)

7 6/67 Syntax of R Reading the set terms (p, r) (p, r) (p, r) (p, r) those who r all p those who r some p those who fail-to-r all p those who r no p those who fail-to-r some p those who don t r some p

8 Syntax of R 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) 7/67

9 8/67 Syntax of R Syntax of R and R 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 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)

10 9/67 Syntax of R Negations We need one last concept, syntactic negation: expression syntax negation 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.

11 10/67 Syntax of R Semantics, again We said before that the semantics was based on [[r]] M M This doesn t give the full semantics of the language, since our syntax now has set terms. We define [[r all p]] = {x M : for all y [[p]], x [[r]] y} [[r some p]] = {x M : for some y [[p]], x [[r]] y} For set terms involving r, we take [[r]] = (M M) \ [[r]], and then [[r all p]] = {x M : for all y semanticsp, x [[r]] y} [[r some p]] = {x M : for some y [[p]], x [[r]] y}

12 11/67 Syntax of R By the way Set terms like (p, r) use the relation in a way that is different from the Kripke semantics for modal logic. [[r all p]] = {x M : for all y [[p]], x [[r]] y} [[ r p]] = {x M : for all y such that x[[r]]y, y [[p]]}

13 12/67 Syntax of R Review of the notation { No x see any y, All z see some y } = No x are z (x, (y, r)), (z, (y, r)) = (x, z) All x see all y, All p are y, Some p are q = All x see some y (x, (y, r)), (p, y), (p, q) = (x, (y, r))

14 13/67 Syntax of R The hardest two families of inferences (x 1, (x 2, r)) (x n, (y, r)) (x 1, (y, r)) (x 1, (y, r)) (H) (y 1, (y 2, r)) (y n, (z, r)) (z, (y 1, r)) (z, x) (x, x) (x, (y 1, r)) (HH) To see that (H) is semantically valid, argue by cases as to whether x 1 or not. For (HH), argue by cases as to whether y 1 or not.

15 14/67 The logical system: examples 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

16 15/67 The logical system: examples 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 RAA and (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.

17 16/67 The logical system: examples Let s see some examples Here is a derivation showing that (x, (y, r)), (p, y), (p, q) (x, (y, r)) (p, q) (p, y) (x, (y, r)) (p, y) (x, (p, r)) (p, y) (x, (y, r))

18 17/67 The logical system: examples Let s see some examples Here is a formal proof showing that (x, x) (y, (x, r)) In words, if there are no xs, then all y s have any relation whatsoever to all of them. Note that this does not follow from the rule (p, p) (p, c)

19 17/67 The logical system: examples Let s see some examples Here is a formal proof showing that (x, x) (y, (x, r)) In words, if there are no xs, then all y s have any relation whatsoever to all of them. Here is a derivation: [ (y, (x, r))] 1 (x, x) (x, x) (x, x) (y, (x, r)) (RAA)1 Note: the general definition of Γ ϕ now must talk about uncancelled leaves in the derivation.

20 18/67 The logical system: examples Example of a proof in this system What do you think? All X see all Y, All X see some Z, All Z see some Y = All X see some Y

21 18/67 The logical system: examples Example of a proof in this system What do you think? All X see all Y, All X see some Z, All Z see some Y The conclusion does indeed follow. We should have a formal proof. = All X see some Y

22 18/67 The logical system: examples Example of a proof in this system What do you think? 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 X All X see some Z Some X see some Z Z X abbreviates Some X are X All Z see some Y Some Z see some Y Y All X see all Y All X see some Y Some X see no Y Some X aren t X

23 19/67 The logical system: examples But now [Some X see no Y ] X This shows that All X see some Z Some X see some Z Z All Z see some Y Some Z see some Y 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 All X see all Y, All X see some Z, All Z see some Y All X see some Y

24 20/67 The logical system: examples An important consequence of RAA: Proof by Cases If Γ, ϕ ψ, and Γ, ϕ ψ, then Γ ψ For the proof, note first that Γ, ϕ, ψ. So by RAA, Γ, ψ ϕ. Take a derivation showing that Γ, ϕ ψ, replace all leaves ϕ with derivations showing Γ, ψ ϕ. In this way, we see that Γ, ψ ψ. Thus Γ, ψ. So as desired, Γ ψ.

25 21/67 The logical system: examples Another important consequence of RAA A set Γ is consistent if Γ. Γ is complete if for all ϕ, either ϕ Γ or ϕ Γ. In any logical system with RAA, every consistent set Γ has a consistent and complete extension. (For the proof, one can either use Zorn s Lemma, or else successively to Γ add each sentence or its negation.)

26 22/67 Negative result on R A negative result on this language R We are going to prove the completeness of the logic for R, but before that, we argue that RAA is needed. Theorem There are no finite syllogistic logical systems which are sound and complete for R. However, there is a logical system (presented above) which uses reductio ad absurdum [ϕ] and which is complete.. (p, p) ϕ RAA

27 23/67 Negative result on R There are no finite, sound, and complete purely syllogistic logics for R Suppose towards a contradiction that did it. We allow rules with arbitrarily many premises. Fix n N greater than the number of premises in any rule in the system. Let Y 1,..., Y n be distinct variables. Let Γ be the following set of R-formulas: All Y i see some Y i+1 (1 i < n) All Y 1 see all Y n All Y i are Y i (1 i < n) All Y i aren t Y j (1 i j n)

28 Negative result on R There are no finite, sound, and complete purely syllogistic logics for R Suppose towards a contradiction that did it. We allow rules with arbitrarily many premises. Fix n N greater than the number of premises in any rule in the system. Let Y 1,..., Y n be distinct variables. Let Γ be the following set of R-formulas: All Y i see some Y i+1 (1 i < n) All Y 1 see all Y n All Y i are Y i (1 i < n) All Y i aren t Y j (1 i j n) Observe that Γ = All Y 1 see some Y n, but this sentence is not in Γ. 23/67

29 24/67 Negative result on R Proof, continued Γ = All Y i see some Y i+1 (1 i < n) All Y 1 see all Y n All Y i are Y i (1 i < n) All Y i aren t Y j (1 i j n) For 1 i < n, let i = Γ \ {All Y i see some Y i+1 }. Claim If ϕ R and i = ϕ, then ϕ Γ.

30 24/67 Negative result on R Proof, continued Γ = All Y i see some Y i+1 (1 i < n) All Y 1 see all Y n All Y i are Y i (1 i < n) All Y i aren t Y j (1 i j n) For 1 i < n, let i = Γ \ {All Y i see some Y i+1 }. Claim If ϕ R and i = ϕ, then ϕ Γ. It follows this claim that Γ All Y 1 see some Y n.

31 Negative result on R Why the claim establishes the result Claim If ϕ R and i = ϕ, then ϕ Γ. Here is why Γ All Y 1 see some Y n. We show by induction on proof trees using that all deductions from Γ must have an element of Γ on the root. No rule of has more than n 1 premises. By induction hypothesis, the sentences just above the root are contained in Γ. So by the claim, the root is in Γ. Therefore the logic is not complete. 25/67

32 26/67 Negative result on R Claim: if ϕ R and i = ϕ, then ϕ Γ. Γ = All Y i see some Y i+1 (1 i < n) All Y 1 see all Y n All Y i are Y i (1 i < n) All Y i aren t Y j (1 i j n) For 1 i < n, let i = Γ \ {All Y i see some Y i+1 }. Proof sketch We consider every sentence in the language R. We check that all are either in Γ or are falsified in some model of i. All Y i are Y j Some Y i are Y j All Y i see all Y j All Y i see some Y j Some Y i see all Y j Some Y i see some Y j All Y i aren t Y j No X are Y j Some Y i aren t Y j All Y i don t see all Y j No X sees any Y j All Y i don t see some Y j No X sees all Y j Some Y i don t see any Y j Some Y i don t see some Y j

33 27/67 Negative result on R The proof goes on y 1 y 2 y i y i+1 y n 1 y n This structure satisfies i and makes a few false: All Y i are Y j No Y j are Y k for j = k Some Y i are Y j Some Y i aren t Y j All Y i r all Y j All Y i don t r all Y j No Y i sees any Y j All Y i r some Y j All Y i don t r some Y j No Y i sees all Y j Some Y i r all Y j Some Y i don t r any Y j Some Y i r some Y j Some Y i don t r some Y j

34 28/67 Negative result on R The proof goes on The empty structure satisfies i and makes a few more false: All Y j are Y k Some Y j are Y k All Y j r all Y k All Y j r some Y k Some Y j r all Y k Some Y j r some Y k No Y j are Y k for j = k Some Y j aren t Y k for j k All Y j don t r all Y k No Y i sees any Y j All Y j don t r some Y k No Y i sees all Y j Some Y j don t r any Y k Some Y j don t r some Y k

35 29/67 Negative result on R What remains We list the remaining sentences. All Y j r all Y j+1 All Y 1 r some Y n All Y j don t r all Y k No Y i sees any Y j All Y j don t r some Y k No Y i sees all Y j We only list the sentences not in i which are true in both of the two models we have seen so far. The ones on the right are easy to falsify in a model of i.

36 30/67 Negative result on R What remains We list the remaining sentences. All Y j r all Y j+1 All Y 1 r some Y n We can also falsify each of these in a model of i.

37 31/67 Negative result on R What to do? We should not be deterred by the negative result: there are two ways to go: 1 Move from a pure syllogistic system to a more liberal type of logic. (We have already done this by adding RAA.) 2 Use infinitely many rules. (I believe this is possible, but it is tedious.)

38 32/67 The logical system R for R 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

39 33/67 The logical system R for R Completeness We shall show that every set Γ of sentences which is consistent in R is satisfiable. By what we saw earlier, we may assume that Γ is R-complete: for every sentence θ of R, either θ or θ belongs to Γ.

40 34/67 The logical system R for R The architecture of the completeness proof Start with a consistent and complete Γ, and build a model M such that for positive sentences ϕ, M = ϕ iff Γ ϕ All p are q (p, q) Some p are q (p, q) All p r all q (p, (q, r)) positive 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)) negative 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))

41 34/67 The logical system R for R The architecture of the completeness proof Start with a consistent and complete Γ, and build a model M such that for positive sentences ϕ, M = ϕ iff Γ ϕ From this, we show that if ϕ is negative and ϕ Γ, we again have M = ϕ. For if not, then ϕ is positive and M = ϕ. So Γ ϕ. And now Γ is inconsistent.

42 The logical system R for R: completeness Model construction For such a consistent and R-complete set Γ, we shall define a model M = M(Γ) as follows: we let M = {x 1, x 2 : Γ (x, x)} {{p, q} : p q and Γ (p, q)}. We assume this union is disjoint, and we call the elements {p, q} pair-elements. So we have two copies of every variable x such that Γ entails the existence of x, and also pair elements {p, q} corresponding to sentences of the form (p, q) which are provable from Γ and such that p q. Our semantics will insure that the pair-element {p, q} belongs to [[p]] [[q]], and so this element will witness the truth of (p, q) in the model which we are constructing. 35/67

43 36/67 The logical system R for R: completeness Model construction The unary atoms x are interpreted in our models as follows: w i [[x]] iff Γ (w, x) {p, q} [[x]] iff Γ (p, x), or Γ (q, x) For the binary atom r, we need a lot more work. First, suppose M contains x 1, x 2, y 1, and y 2.

44 37/67 The logical system R for R: completeness The picture of [[r]] ({x 1, x 2 } {y 1, y 2 }) x 1 y 1 x 2 y 2 (x, (y, r)) x 1 y 1 x 2 y 2 (x, (y, r)) (x, (y, r)) (x, (y, r)) x 1 y 1 x 2 y 2 (x, (y, r)) (x, (y, r)) (x, (y, r)) x 1 y 1 x 2 y 2 (x, (y, r)) (x, (y, r)) (x, (y, r)) x 1 y 1 y 2 x 2 (x, (y, r)) (x, (y, r)) (x, (y, r)) x 1 y 1 x 2 y 2 (x, (y, r)) It depends on which sentences are provable from Γ.

45 38/67 The logical system R for R: completeness The model in full x i [[r]]y j iff x i y j according to the pictures from before {x, y}[[r]]w 2 iff Γ (x, (w, r)) or Γ (x, (y, r)) {x, y}[[r]]w 1 iff {x, y}[[r]]w 2, or Γ (x, (w, r)), or Γ (y, (w, r)) u 1 [[r]]{x, y} iff Γ (u, (x, r)) or Γ (u, (y, r)) u 2 [[r]]{x, y} iff u 1 [[r]]{x, y}, or Γ (u, (x, r)), or Γ (u, (y, r)) {x, y}[[r]]{p, q} iff Γ (x, (p, r)), or Γ (x, (q, r)), or Γ (y, (p, r)), or Γ (y, (q, r))

46 39/67 The logical system R for R: completeness Facts Concerning the relation [[r]]: 1 x 1 [[r]]y 2 iff Γ (x, (y, r)). 2 x 1 [[r]]y 1 iff Γ (x, (y, r)) or Γ (x, (y, r)). 3 x 2 [[r]]y 2 iff Γ (x, (y, r)) or Γ (x, (y, r)). 4 x 2 [[r]]y 1 iff for some i and j, x i [[r]]y j.

47 The logical system R for R: completeness The easy part Let Γ be an arbitrary set of R-sentences. Let ϕ be a positive sentence. If Γ ϕ, then M(Γ) = ϕ. We argue by cases on ϕ. Case 1: ϕ is (x, y). If z i [[x]], then z x. By monotonicity, z y. So z i [[y]]. If {p, q} [[x]], then without loss of generality (p, x). Again, we see that {p, q} [[y]]. Case 2: ϕ is (x, y). This time {x, y} is an element of our model. Our logic contains the identity axioms All x are x. By our semantics, {x, y} [[x]] [[y]]. Thus the model overall satisfies Some x are y. Case 3: ϕ is (x, (y, r)). Let z i [[x]] and w j [[y]]; so we have z x and w y. By monotonicity, Γ (z, (w, r)). So z i [[r]]w j for 1 i, j 2. We also must consider pair-elements {p, q} [[x]]. Without loss of 40/67

48 The logical system R for R: completeness More Case 4: ϕ is (x, (y, r)). In this case, we can assume that [[x]]. That is, Γ (x, x). Then Γ (y, y) as well. We shall show that every element of [[x]] is related to y 1. Let z i [[x]], so that z x. By monotonicity, Γ (z, (y, r)). Then by our Facts, parts (2) and (4), we indeed have z i [[r]]y 1 and z i [[r]]y 2. Further, let {p, q} [[x]]. Without loss of generality, p x. By monotonicity, Γ (p, (y, r)). Then by the definition of [[r]], {p, q}[[r]]y 1. Case 5: ϕ is (x, (y, r)). By rule (I) of our logic, Γ (x, x). Let w j [[y]]. Then Γ (w, y). Hence Γ (x, (w, r)). By our Facts, parts (3) and (4), x 2 [[r]]w 1, and also x 2 [[r]]w 2. We must also consider pair-elements of [[y]]. Let {p, q} [[y]] so that Γ (p, q); and assume Γ (p, y). By monotonicity, Γ (x, (p, r)). By construction, x 2 [[r]]{p, q}. We conclude that x 2 is the required witness to (x, (y, r)). Case 6: ϕ is (x, (y, r)). Here both Γ (x, x) and also (y, y). By Proposition 39, part (4), x 2 [[r]]y 1. So M = (x, (y, r)). 41/67

49 42/67 The logical system R for R: completeness The hard part Let Γ be complete and consistent. Let ϕ be a positive sentence. If M(Γ) = ϕ, then ϕ Γ. We argue by cases on ϕ. In each case, we assume that M(Γ) = ϕ, and we then show Γ ϕ. Since Γ is complete, we indeed have ϕ Γ. One fact which we shall use frequently is that if [[x]] in M(Γ), then Γ (x, x). For if y j [[x]], they by the structure of the model, Γ (y, y) and also (y, x). Similar considerations apply to a pair-element {u, w} [[x]]. Case 1: ϕ is (x, y). We may assume that Γ (x, x); if not, then Γ ϕ using (A). And then the structure of the model easily tells us that Γ ϕ. Case 2: ϕ is (x, y). The argument is very close to what we do concerning (x, (y, r)) in Case 6 below.

50 43/67 The logical system R for R: completeness The hard part, continued Case 3: ϕ is (x, (y, r)). By completeness, either Γ (x, x); or Γ (y, y); or else both Γ (x, x) and Γ (y, y). In the first case, Γ ϕ using the rule (A). In the second case, we show easily that Γ ϕ. In the last case, consider M = M(Γ). By the lemma which we have already seen, M(Γ) = ϕ. In M, x 1 [[x]] and y 2 [[y]]. Since M = ϕ, x 1 [[r]]y 2. By inspection of the model, Γ (x, (y, r)).

51 The logical system R for R: completeness More Case 4: ϕ is (x, (y, r)). As in the previous case, we may assume that Γ (x, x). Consider M = M(Γ). In the model, [[x]] by definition of the model, and because M = ϕ, the same is true of y. In particular, x 1 is related to some element of [[y]]. Say x 1 [[r]]z j where z y. If j = 1, then by Proposition 39, part (2), we Γ (x, (z, r)) or Γ (x, (z, r)). In the first case, we are done by monotonicity. So we shall assume that Γ (x, (z, r)). Since z j belongs to the model Γ (z, z). Therefore Γ (x, (z, r)), and as above we are done. Now if j = 2, then by Proposition 39, part (1), we have Γ (x, (z, r)). Exactly as above, we reason that Γ (x, (y, r)). The last possibility is that x 1 [[r]]{p, q}, where Γ (p, q). Without loss of generality, suppose that Γ (x, (q, r)) and also that Γ (p, y). The derivation in Example?? shows that Γ (x, (y, r)). 44/67

52 45/67 The logical system R for R: completeness Yet more Suppose next that it is x 2 [[x]] which is related by [[r]] to y 2. We have two alternatives: Γ (x, (y, r)) (and we are done); or else Γ (x, (y, r)). In this last case, we have already seen Γ (x, x), and we now have Γ (x, (y, r)). Finally, suppose that {p, q} [[x]] and also that {p, q}[[r]]y 2. Either Γ (p, (y, r)) or Γ (q, (y, r)). Since this pair-element {p, q} belongs to our model, Γ (p, q). So either Γ (p, (y, r)) or Γ (q, (y, r)). But also, either p x or q x. Without loss of generality, p x. By monotonicity, Γ (x, (y, r)).

53 The logical system R for R: completeness Concluding the proof Case 6: ϕ is (x, (y, r)). In our final case, we must have Γ (x, x); also Γ (y, y). Suppose that z 1 [[x]] and w 1 [[y]] are related by [[r]]. Thus z x and w y. By examining the model, Γ (z, (w, r)). Since z 1 belongs to our model, Γ (z, z). Thus Γ (z, (w, r)). And by monotonicity again, Γ (x, (y, r)). Next, suppose that z 1 [[x]] and w 2 [[y]] are related by [[r]]. The work here is quite similar, and we omit all the details. The same goes for the case of z 2 [[x]] and w 1 [[y]], and also for the case of z 2 [[x]] and w 2 [[y]]. There are several more cases, owing to the possibility that the witnesses to (x, (y, r)) might include pair-elements. These are all routine, and we omit these details. This concludes the proof. 46/67

54 47/67 The logical system R for R: completeness Completeness of the logical system For Γ {ϕ} R, Γ = ϕ iff Γ ϕ in R. The soundness is an easy induction on derivations. For the completeness, we need only show that a consistent set Γ is satisfiable. We may assume that Γ is R-complete. Consider M = M(Γ). By what we know M satisfies the positive sentences in Γ. We claim that M satisfies the negative sentences in Γ as well. For suppose that ψ is positive and ψ belongs to Γ. If M = ψ, we would have M = ψ. Also by what we know, Γ ψ. But then Γ is inconsistent, a contradiction. The claim shown, we now see that M = Γ.

55 48/67 The logical system R for R: completeness Incorporating background information Suppose we have a stock of background facts about verbs, such as kissing entails touching This background fact cannot be stated in any of the languages which we have so far studied. Nevertheless, it can be made into a semantic requirement: we would require of a model that [[kiss]] [[touch]]. Even though we cannot state our background fact as an axiom, it does yield rules of inference.

56 49/67 The logical system R for R: completeness Incorporating background information More abstractly, suppose we have a rule like r s (1) and we restrict attention to the models where [[r]] [[s]]. We get rules like (d, (c, r)) (d, (c, s)) (d, (c, r)) (d, (c, s)) (d, (c, r)) (d, (c, s)) (d, (c, r)) (d, (c, s)) We add these to the system R. Proposition The system R together with the rules of a sound and complete logic: Γ ϕ iff Γ = ϕ.

57 50/67 R R expression variables syntax unary atom p, q binary atom 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)

58 51/67 R A stronger negative result on the larger language R Theorem There are no finite, purely syllogistic logical systems which are sound and complete for R. As we now know, there is such a system using RAA. Theorem There are no finite, sound and complete syllogistic logical systems for R, even allowing RAA. (But see our next lecture set for a rejoinder.)

59 52/67 R The Aristotle Boundary FOL Church-Turing FO 2 R adds full N-negation Aristotle S R relational syllogistic S

60 53/67 R Next: relative clauses FOL Church-Turing FO 2 RC adds full N-negation R Aristotle S R RC add relative clauses = relativized quantifiers S

61 54/67 R Inference with relative clauses What do you think about this one? All skunks are mammals All who fear all who respect all skunks fear all who respect all mammals

62 54/67 R Inference with relative clauses It follows, using an interesting antitonicity principle: All skunks are mammals All who respect all mammals respect all skunks

63 54/67 R Inference with relative clauses It follows, using an interesting antitonicity principle: All skunks are mammals All who respect all mammals respect all skunks All who fear all who respect all skunks fear all who respect all mammals

64 55/67 R RC and RC RC allows sentential subjects to be noun phrases containing subject relative clauses. who r all p who don t r all p who r some p who don t r any p expression syntax RC sentence (d +, c) (d +, c) RC sentence (d, c) (d, c) d + is a positive set term, and c is an arbitrary set term.

65 56/67 R The logical system RC (c +, c + ) (T) (c +, d) (c +, c + ) (I) (b +, c + ) (c +, d) (b +, d) (B) (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)) (K) (p, q) ( (p, r), (q, r)) (L) (q, (p, r)) (p, p) (II) (p, p) (c +, (p, r)) (Z)

66 57/67 R Return of the skunks Iterated relative clauses (s, m) ( (m, r), (s, r)) ( ( (s, r), f ), ( (m, r), f ))

67 58/67 R(tr) and RC(tr) Next: comparative adjectives used for inferences involving phrases like bigger than some kitten FO 2 + trans FOL Church-Turing FO 2 RC (tr)!! R RC adds full N-negation Aristotle S R RC RC(tr) tr adds comparatives, requiring transitivity S

68 59/67 R(tr) and RC(tr) Comparative adjectives: R(tr) Every giraffe is taller than every gnu Some gnu is taller than every lion Some lion is taller than some zebra Every giraffe is taller than some zebra We extend R to a language R(tr) by taking a set A of comparative adjective phrases in the base. In the semantics, we would require that for a A, [[a]] must be a transitive relation (in every model M): if [[a]](x, y) and [[a]](y, z), then [[a]](x, z).

69 60/67 R(tr) and RC(tr) The logical system R(tr) for R(tr) (x, (y, r)) (y, (z, r)) (x, (z, r)) (x, (y, r)) (y, (z, r)) (x, (z, r)) (x, (y, r)) (y, (z, r)) (x, (z, r)) (x, (y, r)) (y, (z, r)) (x, (z, r)) (x, (y, r)) (y, (z, r)) (x, (z, r)) (x, (y, r)) (y, (z, r)) (x, (z, r)) (x, (y, r)) (y, (z, r)) (x, (z, r)) (x, (y, r)) (y, (z, r)) (x, (z, r))

70 61/67 R(tr) and RC(tr) R(tr): an example Every giraffe is taller than every gnu Some gnu is taller than every lion Some lion is taller than some zebra Every giraffe is taller than some zebra (giraffe, (gnu, taller)) (gnu, (lion, taller)) (giraffe, (lion, taller)) (lion, (zebra, taller)) (giraffe, (zebra, taller))

71 62/67 R(tr) and RC(tr) A point on the completeness The proof turns out to be very similar to the completeness proof for the logical system for R. Indeed, we only have to check that when we work in the bigger system R(tr), the model that we built earlier is transitive. And luckily, this is true!

72 63/67 R(tr) and RC(tr) Irreflexivity and finiteness The additional requirement also results in the soundness of the following irreflexivity rule (p, (p, r)) (Irr) As weak as this looks, adding it gives a complete system. We also have a finiteness rule: (p, p) (p, (p, r)) (Fin) On top of irreflexivity and transitivity, this gives a complete system.

73 64/67 R(tr) and RC(tr) Comparative adjectives: RC(tr) Recall that RC has subject relative clauses. We again wish to interpret this on structures with transitive relations interpreting adjectives. In the logic, we add a few rules to the system RC for RC: (p, (q, r)) ( (p, r), (q, r)) (p, (q, r)) ( (p, r), (q, r)) (p, (q, r)) ( (p, r), (q, r)) (p, (q, r)) ( (p, r), (q, r)) This system derives R(tr).

74 65/67 R(tr) and RC(tr) Next: relational converses used for inferences relating bigger and smaller FOL FO 2 + trans Church-Turing RC (tr, opp) Aristotle FO 2 S R R RC (tr) RC RC(tr) RC RC(tr, opp) adds full N-negation adds relative clauses opp adds opposites of comparative adjectives S

75 66/67 R(tr) and RC(tr) Converses of transitive relations On top of all the other syllogistic systems we have seen (p, (q, t)) (q, (p, t 1 )) (p, (q, t)) (q, (p, t 1 )) (scope) (p, (q, r 1 )) ( (q, r), (p, r)) ( (p, r 1 ), (q, r)) (p, (q, r)) ( (p, r), (q, r 1 )) (p, (q, r 1 )) ( (p, r), (q, r 1 )) (q, (p, r 1 )) (p, (q, r)) ( (p, r 1 ), (n, r)) (p, (n, r)) ( ) (p, (q, r)) ( (p, r 1 ), (n, r)) (p, (n, r)) (scope): if some p is bigger than all q, then all q are smaller than some p or other. ( ): if every dog is bigger than some hedgehog, and everything smaller than some dog is bigger than some cat, then every dog is bigger than some cat.

76 67/67 R(tr) and RC(tr) Where we are We covered these in this lecture. FOL FO 2 + trans Church-Turing first-order logic FO 2 + R is trans RC (tr, opp) Peano-Frege Aristotle S FO 2 S S R R RC (tr) RC RC(tr) RC RC(tr, opp) 2 variable FO logic adds full N-negation RC(tr) + opposites RC + (transitive) comparative adjs R + relative clauses S + full N-negation R = relational syllogistic S adds p q S: all/some/no p are q