composition and substitution: learning about language from algebra

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Composition and Substitution: Learning about Language from Algebra. Ken Presting University of North Carolina at Chapel Hill. Introduction. Intensional contexts are defined by substitution failure Johnny heard that Venus is the Morning Star Johnny heard that Venus is Venus - PowerPoint PPT Presentation

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Composition and Substitution:Learning about Language

from Algebra

Ken Presting

University of North Carolina at Chapel Hill

Introduction

• Intensional contexts are defined by substitution failure– Johnny heard that Venus is the Morning Star– Johnny heard that Venus is Venus

• Composition accounts for indefinite application of finite knowledge– ‘p and q’ is a sentence– ‘p and q and r’ is a sentence– …

Role of Recursion

• Syntax– Atomic symbols– Combination rules– Closure principle

• Finiteness– Limited symbols, rules– Infinitely many expressions

Compositional Semantics

• The usual:

– Choose assignments to atoms– Forced valuations for molecules

The Two-Element Boolean Algebra

• The Truth Values

• Just two atomic objects: 2BA = {0, 1}

– Disjunction = max(a, b)– Conjunction = min(a, b)– Negation = 1 – a

It’s almost familiar

• Boolean arithmetic– 0 1 = 1– 0 1 = 0

• Boolean algebra– A B = C– (A B) ~C = C ~C– (A B) ~C = 0

A Homomorphism to 2BA

• Take any old function that labels sentences with 0 or 1.

• For example:

– f(S) = 0 – f(PQ) = 1– etc.

A Homomorphism to 2BA

• Ask: Does this function have the ‘distributive’ – a(b + c) = ab + ac– f(S P) = f(S) f(P)

• and ‘commutative’ properties?– ac = ca– f(~S) = ~f(S)

A Homomorphism to 2BA

…is a compositional semantics for propositional calculus

Sentence Diagrams

• Tree diagrams– Binary– Associativity allows n-ary nodes

• (advanced topic: add leaves for empty expression)

Repetition

• Identical Subtrees

– In many sentences, certain letters appear twice or more

• P & Q P

– Sometimes whole expressions recur• (P & R) (P & R)

Reducing the diagram

• Identify like-labeled leaves

• Identify like-labeled nodes

• Form equivalence classes

• Redraw tree as lattice

– (advanced topics: empty expression as zero; quotient)

Set Membership Model

• Mapping sentences to sets– Set of letters = conjunction– Singleton set = negation– Associativity

• And vs. Nand– Naturalness of negation– Failure of associativity

Comparing lattices

• Embeddings

• Homomorphism

Substitution for a Letter

• Single-letter expressions– Every sentence is a substitution-instance

of ‘P’– Substitution for single letters is easy

• Multiple occurrences of a letter

Substitution for Expressions

• What do these sentences have in common?

(P & Q) v ~(P & Q)

(T & S) v ~(T & S)

Subalgebras

• A subalgebra is a subset which follows the same rules as its container

• In our case, that means ‘is also a sentence’

Quotients

• Ignore specfied details

• In our case, treat a subsentence as a letter

Sentences as Functions

In Algebra, formulas map numbers to each other

– F(x) = mx + b

• Sentences map the language to itself

– (P v ~P)(Q) = Q v ~Q

Sentences as Functions

• Mapping the language to itself

– Atomic Sentence letters map L to itself– No other sentence does

• Complex sentences map the language to a subset of itself

Image of a Sentence

• Image = all the substitution-instances

Image of ‘P v ~P’ is:

Q v ~Q

R v ~R

(Q & R) v ~(Q & R)

(P & Q) v ~(P & Q)

Composition of mappings

• Substitute into a substitution-instance

• Start with– P v ~P

• Substitute for P– (Q v R) v ~(Q v R)

• Substitute for R– (Q v (S & T)) v ~(Q v (S & T))

Sentence Fractions

• Here’s a fraction

R (P & Q)

• The numerator is R

• The denominator is (P & Q)

Fractions and Substitution

• ‘Multiply’

(P & Q) v ~(P & Q)

• by the fraction R (P & Q)

• This will be a substitution!

Sentence Arithmetic

Start with

– (P & Q) v ~(P & Q)

Dividing by (P & Q), gives a lattice with a missing label:

– ‘x’ v ~ ‘x’

But R replaces ‘x’ (this step is by fiat)

– R v ~R

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