Hardegree, Formal Semantics, Handout of 8

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1 Hardegree, Formal Semantics, Handout of 8 1. Bound Pronouns Consider the following example. every man's mother respects him In addition to the usual demonstrative reading of he, x { Mx R[m(x), δ] } there is the following reading, x { Mx R[m(x), x] } according to which he is anaphoric to every man. It cannot be duplicating, since he does not mean every man, and it cannot be reflexively-anaphoric, since every man does not fill the subject role. Accordingly, we have yet another class of anaphoric pronouns, which are usually called bound pronouns. As with our treatment of reflexive pronouns, in formalizing our treatment of bound pronouns, we employ the empty pronoun-root e, given categorially as follows. type(e) = trans(e) = Next, we add anaphoric-role-creating morphemes, where the anaphoric roles are encoded by negative integers. The type-theoretical account is given as follows, where α is an anaphoric role (negative integer). type{(α)} = D α D trans{(α)} = λx α :x Notice that the round-brackets are part of the morpheme, just as the square-brackets are part of the case-marking morphemes, which include all integers now. type{[n]} = DD n trans{[n]} = λx:x n With this in mind, let us do an example. 1. Example 1 Jay respects his mother Jay [+1] [ 1] respects ( 1) he 's mother [+2] λx.x 1 λx.x 1 λz 1.z J λx{x 1 x 1 } λz 1.z λx.x 6 λz 1.z 6 λx 6 :M(x) λz 1 :M(z) λx.x 2 λy 2 λx 1 Rxy λz 1 :M(z) 2 J 1 J 1 R[J,M(z)] λz 1 λx 1 R[x,M(z)] Notice that he creates the negative-one role ( 1), which is subsequently filled by Jay [ 1]. Note the distinction between round and square parentheses. Notice also that the entry for respects his mother is

2 Hardegree, Formal Semantics, Handout of 8 λz -1 λx 1 R[x,M(z)] which is a peculiar two-place predicate subcategorizing for one nominative argument, and one negative argument. This basically treats the "form" as: respects 's mother 2. Example 2 Jay respects his mother if he is virtuous Jay [+1] [ 1] [ 2] respects ( 1) he 's mother [+2] if ( 2) he 1 [+1] is-virtuous λx.x 1 λx.x -1 λx.x -2 λx -1.x λx.x 6 λx -2.x λx.x 1 J λx{x 1 x -1 x -2 } λx -1.x 6 λx 6.M(x) λx -2.x 1 λx 1 Vx λx -1.M(x) λx.x 2 λpλq(p Q) λx -2 Vx λy 2 λx 1 Rxy λx -1.M(x) 2 J 1 J -1 J -2 λy -1 λx 1 R[x,M(y)] R[J,M(J)] J -2 λx -2 λq (Vx Q) VJ R[J,M(J)] Notice that each instance of he creates its own role, both being filled by Jay, which additionally fills the nominative-role. Whereas an NP can fill any number of anaphoric-roles, it can only fill one thematic-role. 3. Example 3 Jay respects his mother if she is virtuous Jay [+1] [ 1] respects ( 1) he 's mother [+2] [ 2] if ( 2) she [+1] is-virtuous λx.x 1 λx.x -1 λx -1.x λx.x 6 λx.x 2 λx.x -2 λx -2.x λx.x 1 J λx{x 1 x -1 } λx -1.x 6 λx 6.M(x) λx -2.x 1 λx 1 Vx λx -1.M(x) λx{x 2 x -2 } λpλq(p Q) λx -2 Vx λy 2 λx 1 Rxy λx -1 { M(x) 2 M(x) -2 } J 1 J -1 λy -1 { λx 1 R[x, M(y)] M(y) -2 } R[x, M(J)] M(J) -2 λx -2 λq (Vx Q) V[M(J)] R[x, M(J)] In the previous examples, we see how items of type D can bind pronouns. In the following examples, we see how quantifier-phrases can do this equally well. 1 Henceforth, we ignore the empty component of a pronoun, and simply write it already attached to a rolecreating morpheme, which we place in front of the pronoun. We continue to spell the pronoun-root using proper number and gender, but we use the nominative case he, she, it, they. In some cases, the resulting spelling is strange sounding, but not completely absurd heem heez.

3 Hardegree, Formal Semantics, Handout of 8 4. Example 4 every man [ 1] respects ( 1) his mother The negative-case markers indicate that he is bound by every man. The following is the corresponding semantic tree. By convention, we place the "forward looking" marker after the noun phrase, and we place the "backward looking" marker in front of the pronoun. every man [+1] [ 1] respects ( 1) he 's mother [+2] λp 0 {x Px} M 0 λx.x 1 λx.x -1 λx -1.x λx.x 6 { x Mx } λx{x 1 x -1 } λx -1.x 8 λx 6 :M(x) λx -1 :M(x) λx.x 2 λy 2 λx 1 Rxy λx -1 :M(x) 2 { x 1 x -1 Mx } λy -1 λx 1 R[x,M(y)] { R[x,M(x)] Mx } x { Mx R[x,M(x)] } The previous example can also be analyzed using reflexive pronouns. The following one, however, cannot be done this way. 5. Example 5 every man's father respects him We cannot read he as duplicative, since he does not duplicate every man ; if it did, the sentence would mean: every man's father respects every man Also, we cannot read him as reflexive, since its antecedent is not the subject of the verb; if it were, the sentence would mean: every man's father respects himself Rather, the correct analysis treats he as bound by every man, as follows. every man 's [ 1] father [+1] respects ( 1) he m [+2] λp 0 {x Px} M 0 λx.x 6 λx.x -1 λx -1.x λx.x 2 { x Mx } λx{x 6 x -1 } λy 2 λx 1 Rxy λx -1.x 2 { x 6 x -1 Mx } λx 6 :F(x) { F(x) x -1 Mx } λx.x 1 { F(x) 1 x -1 Mx } λy -1 λx 1 Rxy { R[F(x),x] Mx } x { Mx R[F(x),x] }

4 Hardegree, Formal Semantics, Handout of 8 6. Example 6 every man whose mother respects his father respects his father every 1 man whose mother respects his father [+1] [ 2] respects ( 2) he 's father [+2] λp 0 {x Px} λx 0 {Mx & R[M(x), F(x)]} λx.x 1 λx.x -2 λx -2.x λx.x 6 { x Mx & R[M(x), F(x)] } λx{x 1 x -2 } λx -2.x 6 λx 6 :F(x) λx -2 :F(x) λx.x 2 λy 2 λx 1 Rxy λx -2 :F(x) 2 { x 1 x -2 Mx & R[M(x), F(x)] } λy -2 λx 1 R[x, F(y)] { R[x, F(x)] Mx & R[M(x), F(x)] } x { {Mx & R[M(x), F(x)] } R[x, F(x)] } In the above tree, the second he could be treated as reflexive, but not the first he, whose computation appears in the following. man who se [ 1] mother [+1] respects ( 1) he 's father [+2] λx.x 6 λx.x -1 λx -1.x λx.x 6 λx 0.x λx{x 6 x -1 } λx -1.x 6 λx 6 :F(x) λx 0 {x 6 x -1 } λx 6 :M(x) λx -1 :F(x) λx.x 2 λx 0 { M(x) x -1 } λx.x 1 λy 2 λx 1 Rxy λx -1 :F(x) 2 λx 0 { M(x) 1 x -1 } λy -1 λx 1 R[x, F(y)] M 0 λx 0 R[M(x), F(x)] 1 λx 0 { Mx & R[M(x), F(x)] } 7. Example 7 every man is happy if he is virtuous every man [+1 [ 1] is happy if ( 1) he [+1] is virtuous λp 0 {x Px} M 0 λx:x 1 λx:x -1 λx -1 :x λx:x 1 { x Mx } λx{x 1 x -1 } λx -1 :x 1 λx 1 Vx { x 1 x -1 Mx } λx 1 Hx λpλq{pq} λx -1 Vx { Hx x -1 Mx } λx -1 λq {VxQ} { VxHx Mx } x { Mx (Vx Hx) }

5 Hardegree, Formal Semantics, Handout of 8 In this example, we employ a "robust" if-then connective, symbolized by bold-arrow (). Its precise logic is left open, except that it is stronger than [i.e., PQ P Q]. The following example indicates that the converse entailment fails [i.e., P QPQ]. 8. Example 8 no man is happy if he is virtuous no man [+1 [ 1] is happy if ( 1) he [+1] is virtuous λp 0 {x Px} M 0 λx:x 1 λx:x -1 λx -1 :x λx:x 1 { x Mx } λx{x 1 x -1 } λx -1 :x 1 λx 1 Vx { x 1 x -1 Mx } λx 1 Hx λpλq{pq} λx -1 Vx { Hx x -1 Mx } λx -1 λq {VxQ} { VxHx Mx } ~ x{ Mx & (Vx Hx) } x{ Mx ~(Vx Hx) } x{ Mx (Vx ~Hx) } Notice in particular that, if we replace by, then since (P Q) P& Q, we end up with the following translation. x{ Mx (Vx & Hx) } every man is virtuous but not happy Needless to say, this is not what we want. On the other hand, if we adopt a conditionalconnective that satisfies the following condition 2 namely, (PQ) P Q then we get exactly what we want. 9. Example 9 every man respects some woman who respects him every man [+1] [ 1] respects some woman who [+1] respects ( 1) he m[+2] [+2] λp 0 {x Px} M 0 λx.x 1 λx.x -1 λx 0.x λx.x 1 λz -1.z λx.x 2 { x Mx } λx{x 1 x -1 } λy 2 λx 1 Rxy λz -1.z 2 λx 0.x 1 λz -1 λx 1 Rxz W 0 λz -1 λx 0 Rxz λp 0 {y Py} λz -1 λx 0 {Wx & Rxz} λz -1 { y Wy & Ryz } λx.x 2 λy 2 λx 1 Rxy λz -1 { y 2 Wy & Ryz } { x 1 x -1 Mx } λz -1 { λx 1 Rxy Wy & Ryz } { { Rxy Wy & Ryx } Mx } x { Mx y{wy & Ryx & Rxy} } 2 Originally due to Stalnaker in his account of counterfactual conditionals.

6 Hardegree, Formal Semantics, Handout of 8 This computation grants wide-scope to every man. The following computation grants widescope to some woman who. { x 1 x -1 Mx } λz -1 { λx 1 Rxy Wy & Ryz } λz -1 { { Rxy x -1 Mx } Wy & Ryz } Observe that this cannot compute to a truth-value, so it is a bogus computation. This indicates that if every man binds him, then some woman who cannot have wide scope. On the other hand, if we drop the premise that every man binds him, then we obtain the following. { x 1 Mx } λz -1 { λx 1 Rxy Wy & Ryz } λz -1 { { Rxy Mx } Wy & Ryz } λz -1 y{ Wy & Ryz & x{mx Rxy} } This understands him as anaphoric to an antecedent phrase in the surrounding discourse. 10. Example 10 no man respects any woman who respects him no man [+1] [ 1] respects any woman who [+1] respects ( 1)he m[+2] [+2] λp 0 {x Px} M 0 λx.x 1 λx.x -1 λx 0.x λx.x 1 λz -1.z λx.x 2 { x Mx } λx{x 1 x -1 } λy 2 λx 1 Rxy λz -1.z 2 λx 0.x 1 λz -1 λx 1 Rxz W 0 λz -1 λx 0 Rxz λp 0 {y Py} λz -1 λx 0 {Wx & Rxz} λz -1 { y Wy & Ryz } λx.x 2 λy 2 λx 1 Rxy λz -1 { y 2 Wy & Ryz } { x 1 x -1 Mx } λz -1 { λx 1 Rxy Wy & Ryz } { { Rxy Wy & Ryx } Mx } { Rxy Mx & Wy & Ryx } 3 x y{ Mx & Wy & Ryx & Rxy } 3 This is a further variant of the compositional rule for no-any.

7 Hardegree, Formal Semantics, Handout of 8 2. Pronoun-Fronting The syntactic phenomenon known as "fronting" involves taking an interior phrase and moving it to the beginning of a sentence as a way of emphasizing it. This is common in Yiddish-English, and it is common in the speech of the character Yoda in Star Wars. 4 A similar phenomenon occurs when a subject expression is repeated by way of an anaphoric pronoun, as in: Jay, he is virtuous Jay [ 1] ( 1) he [+1] is virtuous J λx.x -1 λx -1 :x λx.x 1 λx -1 :x 1 λx 1 Vx J -1 λx -1 Vx VJ In this sentence, Jay does not fill the subject-role directly, but only indirectly, by being the antecedent of he, which does fill the subject-role directly. The fronting-procedure can be utilized as a way of understanding quantifier phrases, so that (for example) every man respects his mother can be analyzed as having a tacit (unpronounced) occurrence of he, as follows. 5 every man [, he] respects his mother every man [ 1] [ ( 1) he ] [+1] [ 2] respects ( 2) he 's mother [+2] { x Mx } λx.x -1 λx.x 1 λx.x -2 λx -2 :x λx.x 6 λx -1 :x λx{x 1 x -2 } λx -2 :x 6 λx 6 :M(x) λx -1 {x 1 x -2 } λx -2 :M(x) λx.x 2 λy 2 λx 1 Rxy λx -2 :M(x) 2 { x -1 Mx } λx -1 R[x, M(x)] { R[x, M(x)] Mx } x { Mx R[x, M(x)] } λy -2 λx R[x, M(y)] Notice that, according to this analysis, the second he is anaphoric to the first he, which is anaphoric to every man. 4 It is accordingly often called Y-fronting [presumably for Yiddish, not Yoda!] 5 This is formally parallel to the transformation quantifier-raising.

8 Hardegree, Formal Semantics, Handout of 8 A more complex example is the following. every man [, he] 's mother respects his father every man [ 1] [ ( 1) he 's [ 2] mother [+1] respects ( 2) he 's father [+2] λx.x 6 λx.x -2 λy 2 λx 1 Rxy λx -2 :x λx.x 6 λx 6 :M(x) λx.x 2 λx -1 :x λx{x 6 x -2 } λx -1 {x 6 x -2 } λx 6 :M(x) λx -1 {M(x) x -2 } λx.x 1 λx -1 {M(x) 2 x -2 } λy -2 λx R[x, F(y)] { x -1 Mx } λx -1 R[M(x), F(x)] { R[M(x), F(x)] Mx } x { Mx R[M(x), F(x)] } 3. Reflexive Pronouns Revised Now that we have bound pronouns, and role-creation, we revise our treatment of reflexion, as follows. ref(θ,α) = λx θ {x θ x α } Here, as usual, θ is a case-marker [non-negative integer 6 ], and α is an anaphoric-marker [negative integer]. In the following example, θ=1 and α= 1. Jay [+1] ref(1, 1) respects ( 1) he 's mother [+2] J λx.x 1 λx 1 {x 1 x -1 } λy -1 λx 1 R[x, M(y)] J 1 λx 1 R[x,M(x)] R[J, M(J)] Next, notice what happens if we re-parse the sentence so that [ref(1, 1)] attaches to Jay. Jay [+1] ref(1, 1) respects ( 1) he 's mother [+2] λx.x 1 λx 1 {x 1 x -1 } J λx{x 1 x -1 } J 1 J -1 λy -1 λx 1 R[x, M(y)] R[J, M(J)] The latter looks a lot like our treatment of he as bound by Jay. The latter is not suggested as a proper syntactic analysis, but simply to exhibit formal similarities between reflexive pronouns and bound pronouns. 7 6 Note that the nullative case-marker is included here, since it is needed for common nouns modified by bare adjectives containing a reflexive but not containing who. 7 This parsing could be made semantically inadmissible, by semantically rendering reflexion as follows. ref = λ:λy α λx θ Txy:λx θ Txx This basically takes a transitive verb [two-place predicate] and produces an associated intransitive verb [oneplace predicate].