Building Meaning With The Lambda Calculus

One way to generate the logical form of a sentence, and hence to associate truth-conditions with a sentence, is to use the lambda calculus to represent the semantics of the lexical items in the lexicon and to understand grammatical structure in terms of substitutions in the lambda calculus. (λ is the eleventh letter of the Greek alphabet.)


The Lambda Calculus

Here is a formula in the lambda calculus

λx.MAN(x)

In this formula, the prefix

λx

binds the occurrence of the variable

x

in the one-place predicate

MAN(x)



The Substitution Process

The formula λx.MAN(x) may be understood as a one-place function. The lambda binds a variable that holds a place for a substitution of the argument in the one-place predicate. The symbol @ separates the functor from the argument. In the expression

λx.MAN(x)@vincent

The left hand expression

λx.MAN(x)

is the functor. The right hand expression

vincent

is the argument. The substitution process is called β-reduction. In the example, the result of the conversion is the sentence

MAN(vincent)



Grammatical Categories

This shows that we can think of the semantics of grammatical categories in terms of the lambda calculus. Here are the semantic representations of some lexical entries in the grammar

'every' : λP.λQ.∀x(P@x → Q@x)

'some' or 'a' : λP.λQ.∃x(P@x ∧ Q@x)

'no' : λP.λQ.∀x(P@x → ¬Q@x)

'boxer' : λy.BOXER(y)

'walks' : λx.WALKS(x)


An example helps show how this works. Here is the representation tree for the noun phrase 'every boxer'

   
                every boxer (NP)
      λP.λQ.∀x(P@x → Q@x)@λy.BOXER(y)
                /               \
               /                 \
              /                   \                             
      every (DET)             boxer (NOUN)
      λP.λQ.∀x(P@x → Q@x)     λy.BOXER(y)     
      

The expression at the root of the tree is

'every boxer'
λP.λQ.∀x(P@x → Q@x)@λy.BOXER(y)

By substitution, this reduces to

'every boxer' (NP)
λQ.∀x(λy.BOXER(y)@x → Q@x)

This reduces to

'every boxer' (NP)
λQ.∀x(BOXER(x) → Q@x)

If this noun phrase is given a verb phrase, the result is a sentence. Here, for example, is the representation tree (with substitutions completed) for the sentence 'every boxer walks'

      
                      every boxer walks (S)
                      ∀x(BOXER(x) → WALKS(x))
                      /                 \
                     /                   \                       
                    /                     \                       
                   /                       \      
           every boxer (NP)              walks(VP)
           λQ.∀x(BOXER(x) → Q@x)         λx.WALKS(x)
                /               \
               /                 \
              /                   \                             
          every (DET)             boxer (NOUN)
          λP.λQ.∀x(P@x → Q@x)     λy.BOXER(y)
      
      

Here is the representation tree for the sentence 'vincent loves mia'

     
                    vincent loves mia (S) 
                    LOVE(vincent, mia)  
                      /               \
                     /                 \               
                    /                   \
           vincent (NP)             loves mia (VP)
           λP.P@vincent             λz(LOVE(z,mia))
                                /                  \             
                               /                    \
                              /                      \
                          loves (TV)                mia (NP)
                          λX.λz.(X@λx.LOVE)(z,x))   λP.P@mia
      


Types

It is sometimes useful to think of lambda substitution in terms of types. There are two basic types, e and t. The first is the type of entities in the domain of the model. The second is the type for truth values (true and false). Compound types are built from these two basic types.

One place predicates, for example, have the type <e, t>. When a one-place predicate is given an expression of type e, it returns an expression with type t. To understand more clearly how this works, think again about the lambda expression

λx.MAN(x)@vincent

By β-conversion, this expression reduces to

MAN(vincent)

In terms of types, the lambda expression

      
      λx.MAN(x)@vincent
      <e, t>    e
      
reduces to
      
      MAN(vincent)
      t
      
      


A Simple Grammar to Illustrate Types and Type Conversion

Here is a simple grammar whose lexical entries are associated with formulas in the lambda calculus

S → VP NP
NP → NAME
NP → DET N
VP → IV
VP → TV NP

NAME → mia: λP.P@mia
NAME → vincent: λP.P@vincent

DET → every: λP.λQ.∀x(P@x → Q@x)
DET → some: λP.λQ.∃x(P@x ∧ Q@x)

N → man: λx.MAN(x)
N → woman: λx.WOMAN(x)

IV → laughs: λx.LAUGHS(x)

TV → loves: λX.λz.(X@λx.LOVE)(z,x))


Here are the abstract representation trees showing the type conversions

      
      
                         S : t
                       /       \  
                      /         \
                     /           \   
                    /             \
           NP : (e → t) → t      VP : e → t
      

      
                  NP : (e → t) → t       
                           |
                           |
                  NAME : (e → t) → t

      
      
                 NP : (e → t) → t  
                  /          \
                 /            \ 
                /              \
               /                \
              /                  \
DET : (e → t)  → (e → t) → t         N : e → t
      
      
      
                     VP : e → t
                          |
                          |
                     IV : e → t 
                  
                  
                  
                      VP : e → t
                      /       \
                     /         \
                    /           \
         TV : e → (e → t)     NP : (e → t) → t  
      
      
      

Here is the representation tree (with types and lambda formulas) for a specific grammatical sentence, 'every man laughs'

      
      
                      every man laughs (S)
                      ∀x(MAN(x) → LAUGHS(x))
                               t
                      /                 \
                     /                   \                       
                    /                     \                       
                   /                       \      
           every man (NP)              laughs (VP)
           λQ.∀x(BOXER(x) → Q@x)       λx.LAUGHS(x)
           (e→t)→t                     e→t
                /               \
               /                 \
              /                   \                             
          every (DET)             man (NOUN)
          λP.λQ.∀x(P@x → Q@x)     λy.MAN(y)
          (e →t)→(e→t)→t          e→t
      
      



Problems with the Semantics

This semantics is not without its problems. Here are two of the more famous problems.

(Discourse Representation Theory (DRT) is a possible solution to these problems. This theory, however, is beyond the scope of the course.)


1. Indefinite NPs in Donkey sentences

One problem concerns the truth conditions for donkey sentences.

(When Peter Geach first resented the phenomenon (now known as "donkey anaphora"), he used sentences about farmers and donkeys to construct his counterexample.
The word anaphora comes from the Greek word ἀναφορά, meaning "carrying back.")

The truth-conditions for the following sentences with the indefinite NP a donkey

If John owns a donkey, he feeds it.

Every farmer who owns a donkey feeds it.

are

∀x((donkey(x) ∧ own(john, x)) → feeds(john,x))

∀x∀y((farmer(x) ∧ donkey(y) ∧ own(x, y)) → feeds(x, y))

In the first sentence, a donkey (located in the antecedent of a conditional) is a universally quantified expression with wide scope over the material conditional. In the second sentence, a donkey (located in relative clause that modifies a universally quantified NP) is also a universally quantified expression taking wide scope.

So in neither sentence is the indefinite NP an existentially quantified expression.


2. Indefinite NPs in Discourse

Another problem concerns the way indefinite NPs function in discourse. The sentence A dog came in can be part of a discourse

A dog came in. It sat down.
∃x(dog(x) ∧ came_in(x)). sat_down(x).

but the translation does not make sense. The existential quantifier does not bind the variable in the predicate sat_down(x).


Here is another argument in terms of discourse against the semantics.

In classical logic, ∃xφ and ¬∀¬xφ are logically equivalent. ¬(φ ∧ ψ) and φ → ¬ψ are also logically equivalent. So

∃x(dog(x) ∧ came_in(x))

¬∀x(dog(x) → ¬came_in(x))

are logically equivalent, but the sentences

A dog came in.

Not every dog failed to come in.

are not interchangeable. Only the first can be extended in conversation

A dog came in. It sat down.

Not every dog failed to come in. It sat down.


What we accomplished in this lecture

We looked at how to use the lambda calculus to generate the logical form of a sentence.








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