University:
Boston University
Course:
CAS LX 250 | Introduction to Linguistics
Academic year:

2023
Views:

311

Pages:

3
Author:

SpaceCadet101

Calculating Truth Conditions, Node-by-Node Calculate the denotation of each node in the following tree: S VP NP N Mabel V NP hugged N Solly [[ hugged ]] = λy∈De . [λx∈De . x hugged y] (by TN) [[ [V hugged] ]] = [[ hugged ]] = λy∈De . [λx∈De . x hugged y] (by NN) [[ Solly ]] = Solly (by TN) [[ [N Solly] ]] = [[ Solly ]] = Solly (by NN) [[ [NP [N Solly]] ]] = [[ [N Solly] ]] = Solly (by NN) [[ [VP [V hugged] [NP Solly]] ]] = [[ [V hugged] ]] ([[ [NP Solly] ]]) = [λy∈De . [λx∈De . x hugged y]](Solly) = λx∈De . x hugged Solly [[ Mabel ]] = Mabel (by TN) [[ [N Mabel] ]] = [[ Mabel ]] = Mabel (by NN) [[ [NP [N Mabel]] ]] = [[ [N Mabel] ]] = Mabel (by NN) (by FA) [[ [S [NP Mabel] [VP hugged Solly]] ]] = [[ [VP hugged Solly] ]]([[ [NP Mabel] ]]) = [λx∈De . x hugged Solly](Mabel) (by FA) = 1 iff Mabel hugged Solly Question: why do we have: [[ hugged ]] = λy∈De . [λx∈De . x hugged y] instead of: [[ hugged ]] = λy∈De . [λx∈De . y hugged x] ? Remember that when a λ-expression itself contains another λ-expression, it is the leftmost/outermost occurrence of ‘λ’ that introduces the argument variable for the entire, complex λ-expression. This means that in the above λ-expressions, the variable y represents the argument for the entire function denoted by hugged. Since hugged is a transitive verb, it combines syntactically with its direct object (the hugged one), and so it is the direct object that will provide the argument for the function denoted by hugged. This is why the variable y must correspond to the direct object, not the subject. When we turn to the denotation of the VP: [[ hugged Solly ]] = λx∈De . x hugged Solly we find confirmation for the above reasoning. In this λ-expression, the variable x represents the argument for the entire function denoted by hugged Solly. Since hugged Solly is a VP, it combines syntactically with the subject NP (the hugger), and so it is the subject that will provide the argument for the function denoted by hugged Solly. This is why the variably x must correspond to the subject, not the direct object. Question: Why do we put brackets around a λ-expression when applying it to an argument? Compare: (1) [λy∈De . [λx∈De . x hugged y]](Solly)(Mabel) = [λx∈De . x hugged Solly](Mabel) = 1 iff Mabel hugged Solly vs. (2) [λy∈De . [λx∈De . x hugged y](Solly) ](Mabel) = [λy∈ De . Solly hugged y](Mabel) = 1 iff Solly hugged Mabel When a λ-expression itself contains another λ-expression, we need the brackets to indicate which function applies to which argument. In (1), the function [λy∈De . [λx∈De . x hugged y]] is applied to Solly to yield [λx∈De . x hugged Solly]. This latter function is then applied to Mabel. In (2), the function [λx∈De . x hugged y] is applied to Solly to yield [λy∈ De . Solly hugged y] for the larger λ-expression. This latter function is then applied to Mabel. The resulting truth conditions in (1) and (2) are different. If we had omitted the brackets, as in (3): (3) λy∈De . λx∈De . x hugged y (Solly)(Mabel) it would not be clear which truth conditions were intended.