1/37 Adding Some Larry Moss Indiana University Nordic Logic School August 7-11, 2017
2/37 What this chapter is about We started with the very small logic A of All x are y. And most recently we added verbs and relative clauses. At this point, we go back to A and add Some x are y. Tomorrow, we ll want to add some to the language of verbs, and then negation.
3/37 A course first-order logic R S A(RC) all+some all+no propositional logic A
A course first-order logic R S A(RC) all+some all+no propositional logic A My plan is to do a pretty quick treatment of the logic all + some 3/37
4/37 Syntax We start with a set P of unary atoms, also called nouns. The sentences today are All p are q and Some p are q, where p and q are atoms.
5/37 Semantics For the semantics, we use models M that consist of a set M with interpretations [[p]] of the atoms. Then we define M = All p are q iff [[p]] [[q]] M = Some p are q iff [[p]] [[q]]
6/37 Example M = {1,2,3,4,5,6,7,8,9} [[p]] = {1,3,4} [[q]] = {4} [[n]] = {1,7}. [[x]] = Then we have the following facts: M = Some n are p M = Some p are n M = Some p are p M = Some x are x
7/37 Semantic Consequences What should Γ = ϕ mean?
8/37 Another All p are v,all v are q,some v are w = Some p are p. Can you find a counter-model for this? That is, can you find a model where the assumptions are true and the conclusion is false?
9/37 The logic of All and Some All p are p axiom All p are n All n are q All p are q barbara Some p are q Some p are p some 1 Some p are q Some q are p some 2 All q are n Some p are q Some p are n darii
10/37 Example of a derivation in S All n are q All n are p Some n are n darii Some n are p Some p are n some 1 darii Some p are q In words: if there is an n, and if all ns are ps and also qs, then some p is a q.
11/37 A question to provoke your thoughts Γ = All a are b, All a are c, All b are c, All c are b, All c are d, All b are e, All d are g, All f are g, All g are f, All e are h, All b are I, Some d are e, Some I are g f,g h d I e 8 888888 b,c a Does Γ = Some f are h? Does Γ = Some d are I?
12/37 Let s see how the counter-models work Γ = All a are b, All a are c, All b are c, All c are b, All c are d, All b are e, All d are g, All f are g, All g are f, All e are h, All b are I, ϕ: Some d are e ψ: Some I are g f,g h d I e 8 888888 b,c [[a]] = [[b]] = [[c]] = [[d]] = {ϕ} [[e]] = {ϕ} [[f]] = {ϕ,ψ} [[g]] = {ϕ,ψ} [[h]] = {ϕ} [[I]] = {ψ} a Here is a model which shows Γ = Some d are I. M = {ϕ,ψ}. The points of the model are sentences! For all x, put ϕ in [[x]] provided d x or e x. For all x, put ψ in [[x]] provided I x or g x.
13/37 Completeness: if Γ = ϕ, then Γ ϕ When ϕ is an All-sentence, it s just like what we saw in Tuesday s homework. Let s do this when ϕ is a Some-sentence, say Some y are z. So we assume that Γ = Some y are z, and we show using a canonical model that we devise based on our example above that Γ Some y are z.
14/37 Completeness: if Γ = Some y are z, then Γ Some y are z At this point, I want to remind you of some notation from earlier in the course. If we have a set Γ in mind, we sometimes write x y to mean Γ All x are y We do this to save a little space.
15/37 Completeness: if Γ = Some y are z, then Γ Some y are z Let M be the set of all Some-sentences in Γ. We will use letters like ϕ and ψ for these, and we speak of the first and second atoms in these sentences. Here is how we interpret nouns in our model: [[u]] = {ϕ : }
Completeness: if Γ = Some y are z, then Γ Some y are z Let M be the set of all Some-sentences in Γ. We will use letters like ϕ and ψ for these, and we speak of the first and second atoms in these sentences. Here is how we interpret nouns in our model: [[u]] = {ϕ M : one of the two atoms in ϕ is Γ u} = { Some x are y in Γ : either Γ All x are u or Γ All y are u} Lemma M = Γ. Once this is done, we have M = Some y are z. Why? And we use this last fact to show that Γ Some y are z. 15/37
15/37 Completeness: if Γ = Some y are z, then Γ Some y are z Let M be the set of all Some-sentences in Γ. We will use letters like ϕ and ψ for these, and we speak of the first and second atoms in these sentences. Here is how we interpret nouns in our model: Lemma M = Γ. Once this is done, we have M = Some y are z. Why? And we use this last fact to show that Γ Some y are z.
16/37 Completeness: if Γ = Some y are z, then Γ Some y are z Lemma M = Γ. Proof. (first half) Take an All-sentence in Γ, say All a are b. Let ϕ [[a]]. We show that ϕ [[b]]. Suppose that ϕ is Some m are n. Then either m a, or else n a. Since a b, we see that either m b, or else n b. And this means that ϕ [[b]]. That is, M = All a are b.
17/37 Completeness: if Γ = Some y are z, then Γ Some y are z Lemma M = Γ. Proof. (second half) Take a Some-sentence in Γ, say Some a are b. This sentence itself belongs to [[a]] [[b]]. So M = Some a are b.
18/37 Recall that we are assuming that Γ = Some y are z, and then proving that Γ Some y are z. We have a model M, and we showed that M = Γ. So M = Some y are z. Thus, in our model [[y]] [[z]]. Let ϕ Γ belong to [[y]] [[z]]. Let s write ϕ as Some a are b. Since ϕ [[y]], either a y or b y. Since ϕ [[z]], either a z or b z. Finishing completeness
19/37 Finishing completeness Recall that we are assuming that Γ = Some y are z, and then proving that Γ Some y are z. Γ contains the sentence Some a are b. Case 1: a y and a z Case 2: a y and b z Case 3: b y and a z Case 4: b y and b z
20/37 Recall that we are assuming that Γ = Some y are z, and then proving that Γ Some y are z. Γ contains the sentence Some a are b. Case 1: a y and a z Finishing completeness. Some a are b All a are y Some a are a some 1 darii. Some a are y All a are z Some y are a some 2 darii Some y are z Case 2: a y and b z Case 3: b y and a z Case 4: b y and b z
21/37 Recall that we are assuming that Γ = Some y are z, and then proving that Γ Some y are z. Γ contains the sentence Some a are b. Case 1: a y and a z Finishing completeness Case 2: a y and b z Case 3: b y and a z Case 4: b y and b z Some a are b. Some b are a some 2 All b are y Some b are b some 1 darii. Some b are y All b are z Some y are b some 2 darii Some y are z
22/37 Finishing completeness Recall that we are assuming that Γ = Some y are z, and then proving that Γ Some y are z. Case 1: a y and a z Case 2: a y and b z You try this one. Case 3: a z and b y You try this one. Case 4: a z and b z
23/37 To see if Γ Some a are b Construct the All-graph of Γ. Look at All the Some sentences in Γ one at a time. Let s say that one of them is Some u are v. Ask if one of the following 4 conditions hold: We can get from u to a, and from u to b. We can get from u to a, and from v to b. We can get from v to a, and from u to b. We can get from v to a, and from v to b. If this happens for any sentence Some u are v in Γ some, then we know that Γ Some a are b. If not, then we want to make a counter-model, say M. We take the universe M to be the set Γ some. Some u are v [[W]] iff either u W or v W.
24/37 Adding Some-sentences to A(RC) Recall that A(RC) is our language with term formers r all t sentence former All t u. We now want to enlarge this by adding sentence former Some t u. (Of course, we will then want to add term formers r all t. But I m not going to go there today. )
25/37 The natural thing to try All p are p axiom All p are n All n are q All p are q barbara Some p are q Some p are p some 1 Some p are q Some q are p some 2 All q are n Some p are q Some p are n darii All x (r all y) All z y All x (r all z) down
26/37 But this doesn t work! Theorem There is no finite sound and complete set of rules of the kind we have been studying for this language.
26/37 But this doesn t work! Theorem There is no finite sound and complete set of rules of the kind we have been studying for this language. Here is a hint of why this is the case. some c d,all a x,all a y,all (r all a) x,all (r all a) y some x y. Let Γ be the set of assumptions on the left. Claim Γ = some x y Can you think about this? Hint Do it by cases depending on whether [[a]] = or not.
27/37 Proof by Cases We adopt a new rule, (cases). [some x x] ϕ [all z 1 (r 1 all x)] [all z k (r k all x)] ϕ ϕ cases
27/37 Proof by Cases We adopt a new rule, (cases). [some x x] ϕ In words, to prove ϕ from Γ, we may Take any variable (=noun) x. [all z 1 (r 1 all x)] [all z k (r k all x)] ϕ ϕ cases Add some x are x to Γ, and prove ϕ. Add any set of sentences all z (r all x) to Γ, and again prove ϕ.
28/37 We show that 1 Γ {some a a} some x y 2 Γ {all c (r all a)} some x y The first is easy from all a x and all a y. The second comes from Proof by Cases: An example all c (r all a) all (r all a) y all c y all c (r all a) all (r all a) x all c x some c x some x y some c d
The resulting system is complete. However, adding proof features like proof by cases and reductio ad absurdum complicates the proof search algorithm. Comment on complexity Indeed, in a strictly syllogistic system, the relation Γ ϕ is in polynomial time, and adding the extra features could give us proof systems for which this relation is co-np-complete. In other words, if we have a complete logic for which the problem Γ ϕ is co-np-complete. then (assuming P NP), lll we cannot hope to find a finite, purely syllogistic proof system. 29/37
30/37 On the other hand For the particular logic that we are studying, term formers r all t sentence former All t u and Some t u the relation Γ ϕ happens to be in polynomial time.
30/37 On the other hand For the particular logic that we are studying, term formers r all t sentence former All t u and Some t u the relation Γ ϕ happens to be in polynomial time. Even more strangely, it s possible to extend the language and then indeed get a purely syllogistic system!
31/37 We add to our current language a new four-place sentence former with semantics When x = y, we write a x b. a b xy M = a b iff [[x]] [[y]] = implies [[a]] [[b]] xy i.e., (Some x y) (All a b) A new symbol
32/37 A finite, complete set of rules Alex Kruckman & LM (2017) All p are p axiom All p are n All n are q All p are q barbara Some p are q Some p are p some 1 Some p are q Some q are p some 2 All q are n Some p are q Some p are n darii All x (r all y) All z y All x (r all z) down a a r 0 xy a b yx a b r 1 xy a b xy a b r 2 x a b b c xy xy a c xy r 3 all a b a b r 4 xy a b xy r all b r all a r 5 xy a x a y xy xy b r all a xy r 6 a x a y xy xy a b xy r 7 a b u x v y uv xy xy a b xy r 8 some a b a x b y xy xy some x y r 9 some x y a b xy r 10
33/37 We add a complement symbol to our nouns and verbs. We always understand x to be the same as x. Relational Syllogistic Logic No embedded relative clauses, and subjects noun phrases must not contain relative clauses.
33/37 We add a complement symbol to our nouns and verbs. We always understand x to be the same as x. Relational Syllogistic Logic No embedded relative clauses, and subjects noun phrases must not contain relative clauses. formal syntax read in English all(p,q) all p are q some(p,q) some p are q all(p,r allq) all p r all q all(p,r someq) all p r some q some(p,r allq) some p r all q some(p,r someq) some p r some q some(p, q) some p aren t q all(p,q) no p are q some(p,r someq) some p don t-r some q some(p,r allq) some p don t-r any q all(p,r someq) all p don t-r some q all(p,r allq) all p don t-r any q
34/37 A model M for R is a set M, together with interpretation functions [[ ]] : P P(M) [[ ]] : R P(M M) Semantics This means that for each unary atom p P, [[p]] M, and for each binary atom r, [[r]] M M. We interpret literals p and r using set complements: [[p]] = M \[[p]] [[r]] = M 2 \[[r]]. We then interpret terms by subsets of M in the following way [[s all p]] = {m M : for all n [[p]], (m,n) [[s]]} [[s some p]] = {m M : for some n [[p]], (m,n) [[s]]} Finally, M = all(p,c) iff [[p]] [[c]] M = some(p,c) iff [[p]] [[c]]
35/37 One unary atom p, and one binary atom s. Example of the semantics A model M might have M = {w,x,y,z}, [[p]] = {w,x,y} and s as shown below: w x In this model, 7 7777777 y z [[p]] = {z} [[s all p]] = [[ssomep]] = M [[ssomep]] = M Here are two R-sentences true in M: some(p,p), and also all(p,s somep).
36/37 Logic some(p, q) all(q, c) some(p, c) (D1) all(p, q) all(q, c) all(p, c) (B) all(p, q) some(p, c) some(q, c) (D2) all(p,p) (T) some(p,c) some(p,p) (I) all(q, c) some(p, c) some(p, q) (D3) all(p, p) all(p,c) zero some(p,tsomeq) some(q, q) (II) all(p,talln) some(q,n) all(p,tsomeq) (all ) some(p,tsomeq) all(q,n) some(p,tsomen) ( ) [ϕ] all(p,tsomeq) all(q,n) all(p,tsomen) (allall). ϕ RAA
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