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Composition and Substitution: Learning about Language from Algebra Ken Presting University of North Carolina at Chapel Hill.

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Presentation on theme: "Composition and Substitution: Learning about Language from Algebra Ken Presting University of North Carolina at Chapel Hill."— Presentation transcript:

1 Composition and Substitution: Learning about Language from Algebra Ken Presting University of North Carolina at Chapel Hill

2 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 –…

3 Role of Recursion Syntax –Atomic symbols –Combination rules –Closure principle Finiteness –Limited symbols, rules –Infinitely many expressions

4 Compositional Semantics The usual: –Choose assignments to atoms –Forced valuations for molecules

5 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

6 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

7 A Homomorphism to 2BA Take any old function that labels sentences with 0 or 1. For example: –f(S) = 0 –f(P  Q) = 1 –etc.

8 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)

9 A Homomorphism to 2BA …is a compositional semantics for propositional calculus

10 Sentence Diagrams Tree diagrams –Binary –Associativity allows n-ary nodes (advanced topic: add leaves for empty expression)

11 Repetition Identical Subtrees –In many sentences, certain letters appear twice or more P & Q  P –Sometimes whole expressions recur (P & R)  (P & R)

12 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)

13 Set Membership Model Mapping sentences to sets –Set of letters = conjunction –Singleton set = negation –Associativity And vs. Nand –Naturalness of negation –Failure of associativity

14 Comparing lattices Embeddings Homomorphism

15 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

16 Substitution for Expressions What do these sentences have in common? (P & Q) v ~(P & Q) (T & S) v ~(T & S)

17 Subalgebras A subalgebra is a subset which follows the same rules as its container In our case, that means ‘is also a sentence’

18 Quotients Ignore specfied details In our case, treat a subsentence as a letter

19 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

20 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

21 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) …

22 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))

23 Sentence Fractions Here’s a fraction R (P & Q) The numerator is R The denominator is (P & Q)

24 Fractions and Substitution ‘Multiply’ (P & Q) v ~(P & Q) by the fraction R (P & Q) This will be a substitution!

25 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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