Logics with Verbs, Relative Clauses,and Comparative Adjectives

Larry Moss, Indiana University

EASLLC 2014

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Where we are

Aristotl

e

Turing

S

S†

R

RC

RC(tr)

RC(tr , opp)R†

RC†

RC†(tr)

RC†(tr , opp)

FOL

FO2 + trans

FO2

† adds full N-negation

R + relative clauses

relational syllogistic

all these are today

R∗ + (transitive)

comparative adjs

done

done

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Preliminary work, done as a set of groupexercises

I am going to introduce this topic in a slow wayby 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 theproofs,and to “get our hands dirty”.

After we do this, I will go more quickly and omit a huge number ofdetails.

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Does the conclusion follow?

All boys are malesAll women see all malesAll 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.

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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 likeEnglish.Sometimes it would be as “see all who are x .”

So we get set terms like

see all boyssee all (see all women)see all (see all (see all women))

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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 likeEnglish.Sometimes it would be as “see all who are x .”

So we get set terms like

see all boyssee all (see all women)see all (see all (see all women))

Later on, we’ll call set terms set terms.

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Syntax and Semantics

For our sentences, we take all expressions

All x y

where x and y are set terms.

In the semantics, we need an inductive definition to interpret theset terms.

[[see all x]] = {m ∈ M : }

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Syntax and Semantics

For our sentences, we take all expressions

All x y

where x and y are set terms.

In the semantics, we need an inductive definition to interpret theset terms.

[[see all x]] = {m ∈ M : }

[[see all x ]] = {m ∈ M : for all n ∈ [[x ]], (m, n) ∈ [[see]]}

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Example of a model

a

bc

d

e f

[[males]] = {a, d , f }

[[females]] = {b, c , e}

[[boys]] = {a, d}

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 )}

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Example of a model

a

bc

d

e f

[[males]] = {a, d , f }

[[females]] = {b, c , e}

[[boys]] = {a, d}

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))]]?

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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 zBarbara

But note that we are using this with x , y , and zas set terms.

But in fact we are missing at least one rule!

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Logic

All x are xaxiom

All x are y All y are z

All x are yBarbara

All y x

All (see all x)(see all y)skunk

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Logic

All x are xaxiom

All x are y All y are z

All x are yBarbara

All y x

All (see all x)(see all y)skunk

Example: All x (see all y),All z y ⊢ All x (see all z)

All x (see all y)

All z y

All (see all y)(see all z)

All x (see all z)

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Does the conclusion follow?

All boys are malesAll who see all women also see all who see all boysAll who see all see all boys – they are all womenAll women see all malesAll who see all boys see all males

In our language, this is

All boys malesAll (see all women) (see all (see all boys))All (see all (see all boys)) womenAll women (see all males)

All (see all boys) (see all males)

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The canonical model Γ of a set of assertionsin 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, becauseif Γ 6⊢ ϕ, then M 6|= ϕ.

(In other words,every sentence which is true in all models of Γ (hence in M)would be provable from Γ.)

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The canonical model Γ of a set of assertionsin 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

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The canonical model Γ of a set of assertionsin 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}

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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 p0 for p, and pn+1 for see all pn;we adopt similar notation for q.Then the canonical model of Γ is shown below:

p0� _

��

p1oooo p2� _

��

oo

~~||||

||||

p3oo p4� _

��

oo

~~||||

||||

· · ·

q0 q1oo ?�

OO``BBBBBBBBq2

oo q3oo oo ?�

OO``BBBBBBBBq4

oo · · ·

Notice that the notation for subset arrows is different than thearrows for see.Notice also that we have omitted the reflexive subset arrows on allof the nodes,and also that the arrows which are shown alternate directions.

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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.

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Solving the problem

Does this follow?

All boys malesAll (see all women) (see all (see all boys))All (see all (see all boys)) womenAll women (see all males)

All (see all boys) (see all males)

We make the part of the canonical model just using the set termsabove, and all the subterms:

boysmenwomen

see all boyssee all mensee all womensee all see all women

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More

see all see all boys

see all boys

see all males see all females

boys

femalesmales

subset

subset subset

subset

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Here is part of the “see” relationThe logic gives us part of the “see” relation in a trivial way

see all see all boys

see all boys

see all males see all females

boys

femalesmales

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Many verbs

What would we do if we had more than one verb?

All women are football playersAll football fans love all football players

All football fans like all women

Does this work?

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Background knowledge

In addition to our assumptions Γ, we can adopt a set∆ of background assumptions which need not be expressible in ourlanguage.

In this case, we might say

∆ = {loves ⊑ likes}

Then we require that our modelsrespect the background assumptions:[[loves]] ⊆ [[likes]].

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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?

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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 Γand the logic previously described,augmented by the inference rules corresponding to thesentences in ∆

And why does it hold?

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Adding transitive verbsall work on R and related systems is joint with Ian Pratt-Hartmann

The next language uses “see” or r as variables for transitive verbs.

All p are qSome p are q

All p see all qAll p see some qSome p see all qSome p see some q

All p aren’t q ≡ No p are qSome p aren’t q

All p don’t see all q ≡ No p sees any qAll p don’t see some q ≡ No p sees all qSome p don’t see any qSome p don’t see some q

The interpretation is the natural one, using the subject wide scopereadings 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

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

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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.

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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 YSome X are X All X see some Z

Some X see some ZSome Z are Z All Z see some Y

Some Z see some YSome Y are Y All X see all Y

All X see some Y Some X see no YSome X aren’t X

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But now

[Some X see no Y ]

Some X are X All X see some ZSome X see some Z

Some Z are Z All Z see some YSome Z see some Y

Some Y are Y All X see all YAll X see some Y [Some X see no Y ]

Some X aren’t XAll 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

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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))

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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 cpositive p ∀(p, r) ∃(p, r)

negative p ∃(p, r) ∀(p, r)

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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

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More on the set terms

To understand how they work, let us exhibit a rendering of thesimplest set terms into more idiomatic English:

∀(boy, see) those who see all boys∃(girl, admire) those who admire some girl(s)∀(boy, see) those who fail-to-see all boys

= those who see no boys= those who don’t see any boys

∃(girl, recognize) those who fail-to-recognize some girl= those who don’t recognize some girl or other

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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 cpositive p ∀(p, r) ∃(p, r)

negative p ∃(p, r) ∀(p, r)

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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, qbinary 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)

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Negations

We need one last concept, syntactic negation:

expression syntax negation

positive set term c p pp 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.

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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]] onthe right:

w

@@@

@@@@

@ x

~~~~~~

~~~~

��

oo

y

OO

oo // zyy

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[[∀(∃(dog, likes), see)]] = {x}.

Since [[cat]] contains x , we have

M |= ∀(∀(∃(dog, likes), see), cat).

That is, in our model it is true that everything which seeseverything bigger than some dog is a cat.

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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, c̄) ∃(p, c)

∃(p, q̄)

∀(p, p̄)

∀(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)

ϕRAA

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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, p̄)

∀(p, c)

∃(p,∃(q, t))

∃(q, q)

∀(q, c̄) ∃(p, c)

∃(p, q̄)(⋆)

∀(p,∀(n, t)) ∃(q, n)

∀(p,∃(q, t))

Of these, (⋆) is the most interesting.

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Example of a proof in the system for R

Every porter recognizes every porterNo 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

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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.

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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.

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RCA

Now let’s add comparative adjectives,interpreted as transitive relations.

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Example of what we want to formalize inRCA

Example

Every hyena is taller than some jackalEverything taller than some jackal is not heavier than any warthog

Everything which is taller than some hyena—is not heavier than any warthog

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Syntax

expression variables syntax

unary atom p, qadjective atom atv atom vbinary 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)

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A few more examples

∀(boy, see) those who see all boys∃(girl, taller) those who are taller than some girl(s)∀(boy, see) those who fail-to-see all boys

= those who see no boys= those who don’t see any boys

∃(girl, see) those who fail-to-see some girl= those who don’t see some girl

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The set terms in this language are arecursive 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 thansomeone who sees no cat.

We should note that the relative clauses which can be obtained inthis way are all “missing the subject”, never “missing the object”.

The language is too poor to express predicates likeλx .all boys see x .

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Semantics

A model (for this language L) is a pair M = (M, [[ ]]), where M isa non-empty set, [[p]] ⊆ M for all p ∈ P, [[r ]] ⊆ M2 for all binaryatoms r ∈ R.The only requirement is that for all adjectives a, [[a]] should be atransitive relation: if a(x , y) and a(y , z), then a(x , z).

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Semantics

Given a model M, we extend the interpretation function [[ ]] to therest of the language by setting

[[p]] = M \ [[p]][[r ]] = M2 \ [[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 ]] 6= ∅

If Γ is a set of formulas, we write M |= Γ if for all ϕ ∈ Γ, M |= ϕ.

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Logic

∀(c , c)(T)

∃(c , d)

∃(c , c)(I)

∀(c , b)

∀(b, c)(C)

∀(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, p̄)

∀(c ,∀(p, r))(Z)

∀(p, p̄)

∃(∀(p, r),∀(p, r))(W)

∀(p,∃(q, a))

∀(∃(p, a),∃(q, a))(tr1)

∀(p,∀(q, a))

∀(∃(p, a),∀(q, a))(tr2)

∃(p,∀(q, a))

∀(∀(p, a),∀(q, a))(tr3)

∃(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 cover set terms, d over positive set terms, r over binary atoms, and a over adjectiveatoms.

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An invalid inference

The following “inference” is invalid:

Every giraffe sees every gnuSome gnu sees every lionSome lion sees some zebraEvery 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 theinterpretation of the verb is the relation indicated by the arrow.

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Reading the rules

(J) If all watches are gold items, then everyone who owns all golditems owns all watches.

(K) If all watches are gold items, then everyone who owns somewatch owns some gold item.

(L) If some watches are gold items, then everyone who owns allwatches 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 everthingbigger than some watch is bigger than some pencil.

(tr2) If all watches are bigger than all pencils, then everthing biggerthan some watch is bigger than all pencils.

(tr3) If some watch is bigger than all pencils, then everthing biggerthan all watches is bigger than all pencils.

(tr4) If some watch is bigger than some pencil, then everthingbigger than all watches is bigger than some pencil.

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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 andmammal have switched positions.

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Return of the jackals and hyenas

Every hyena is taller than some jackalEverything taller than some jackal is not heavier than any warthog

Everything which is taller than some hyenais not heavier than any warthog

∀(hyena,∃(jackal, taller))

∀(∃(hyena, taller),∃(jkl, taller))(tr1)

∀(∃(jkl, taller),∀(warthog, heavier))

∀(∃(hyena, taller),∀(warthog, heavier)))

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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)(D1)

∀(p, q)(RAA)1

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