Download Semantics 1: Exercise Sheets and more Study notes Logic in PDF only on Docsity!
Eberhard Karls Universit¨at T¨ubingen | Seminar f¨ur Sprachwissenschaft (SfS) PS Semantik 1 (WS 2015/16), Prof. Dr. Wolfgang Sternefeld Tutoren: Thorwald Kassemek, Natalie Clarius
Semantics 1:
Exercise Sheets
with (partial) solutions
This is a compilation of all exercise sheets and some additional material for the proseminar
Semantics1 in WS 2015/16.
Most (yet not all) of the exercises come with an exemplary solution, which is to be seen as
one possible approach to solve the assignment rather than the one and only gold standard
solution. I do not guarantee absolute and ultimate correctness!
The PDF file contains clickable bookmarks which lead you directly to the desired exercise
sheet.
Have fun!
N.C.
Eberhard Karls Universit¨at T¨ubingen | Seminar f¨ur Sprachwissenschaft (SfS) PS Semantik 1 (WS 2015/16), Prof. Dr. Wolfgang Sternefeld Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 1:
Revision
Zimmermann & Sternefeld: Ch. 1- Slides ESSLLI: No. 1- Gamut: Vol. 1 Ch. 1- Partee, Meulen & Wall: Pt. A-B Hodges: Ch. 1-
Exercise 1: Statement logic
Proof that the following formulae of propositional logic are equivalent by means of your choice (e.g. tranformation, truth table, natural deduction, tableaux, ...)
(1)
(p ∧ q) ∨ (q → r)
(¬¬p ∧ ¬¬q) ∨ (¬q ∨ r)
r ∨ ¬q
(p ↔ q) ∧ (p ∧ q)
Exercise 2: Predicate logic
Proof that the following formulae of predicate logic are equivalent by means of your choice (e.g. trans- formation, semantic evaluation, natural deduction, tableaux, ...)
(1) ¬∃x
student(x) ∧ ∃y
pizza(y) ∧ ¬like(x, y)
(2) ∀x
student(x) → ∀y
pizza(y) → like(x, y)
(3) ∀x¬
student(x) ∧ ¬∀y¬
pizza(y) ∧ ¬like(x, y)
What do the three of the formulae mean, i.e. expressed in natural langauge?
Exercise 3: Sets
a. Define the set A which consists of all natural numbers, giving the definition in three different ways.
b. Determine the outcome of the following set-theoretic operations:
(1) A ∩ {x ∈ Z : x%2 = 0}^1 (2) A \ {x ∈ Z : x is a prime number} (3) A ∪ {x ∈ Z : x ∈ Z and x < 0 }
Hint: You do not need to spell out an infinite number of elements ;) It is sufficient to state enough elements to make the content of the set clear.
(^1) % is used here as an operator for retrieving the remainder of a division: 5%3 = 2; 22%7 = 1; 8%4 = 0; ...
Eberhard Karls Universit¨at T¨ubingen | Seminar f¨ur Sprachwissenschaft (SfS) PS Semantik 1 (WS 2015/16), Prof. Wolffeld Sternegang Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 2:
Extensions
Zimmermann & Sternefeld: Ch. 4- Slides ESSLLI: No. 2-
Exercise 1: Relational and functional nouns
Assume this little village in Sweden:
Determine the extensions of the following expressions and classify them, i.e. whether they are relational, functional or of a different category.
(1) neighbour-of (2) lives-across-from (3) lives-directly-right-to (4) has-no-left-neighbour
Solution:
(1) neighbour-of : {< P ippi, M ichel >, < M ichel, P ippi >, < M ichel, Kalle >, ...} ⇒ relational (2) lives-across-from:{< P ippi, K.u.J. >, < K.u.J., P ipp >, < M ichel, Ronja >, ...} ⇒ functional (3) lives-directly-right-to: {< P ippi, M ichel >, < M ichel, Kalle >, ...} ⇒ relational (4) has-no-left-neighbour : {P ippi, K.u.J.} ⇒ neither relational nor functional
Exercise 2: Referential expressions
Compute the extensions of the following expressions.
(1) current president of Germany (2) the current president of Germany
Solution: (1) ⟦current president of Germany⟧s = {Joachim Gauck} (2) ⟦the current president of Germany⟧s = Joachim Gauck
Exercise 3: n-place verbs and plugging
a. Let A be a set such that
A = {< a, c >, < a, b >, < c, q >, < d, f >, < a, e >, < y, e >, < x, d >, < q, w >, < a, y >, < w, e >, < q, c >, < x, a >, < s, w >, < q, a >, < y, x >, < c, d >, < e, w >, < a, x >, < q, c >, < y, s >, < w, s >, < a, d, s >, < w, s, x >, < y, s, x >, < a, q, y >, < w, e, d >, < a, w, x >}
Compute the following operations:
(1) AÐ→∗ s (2) AÐ→∗ y (3) AÐ→∗ a (4) A←Ð∗ q (5) A←Ð∗ w (6) A←Ð∗ x
b. Given is the following sentence:
(1) John lends his pen to Mary.
- Calculate the following sets A, B, C, D by use of the plugging operation such that A. A = ⟦lends⟧s B. B = ⟦lends his pen⟧s C. C = ⟦lends his pen to Mary⟧s D. D = ⟦John lends his pen to Mary⟧s
- What is the extension of (1) given that the sentence is false?
Solution: a. Plugging operations with set A:
(1) AÐ→∗ s = {< y >, < w >, < a, d >} (2) AÐ→∗ y = {< a, q >} (3) AÐ→∗ a = {< x >, < q >} (4) A←Ð∗ q = {< w >, < c >, < a >} (5) A←Ð∗ w = {< e >, < s >, < s, x >, < e, d >} (6) A←Ð∗ x = {< d >, < a >, < s >}
Eberhard Karls Universit¨at T¨ubingen | Seminar f¨ur Sprachwissenschaft (SfS) PS Semantik 1 (WS 2015/16), Prof. Wolfgang Feldstern Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 3:
Extensions, Determiners and Quantifiers
Zimmermann & Sternefeld: Ch. 5- Slides ESSLLI: No. 3-4 + Addenda
Exercise 1: Plugging
Given are the following expressions:
⟦Paul ⟧s = p ⟦sleeps⟧s = {< p >, < a >}
Compute the plugging operation
⟦sleeps⟧s ∗ ⟦p⟧s
and determine its meaning in natural language.
Solution:
⟦sleeps⟧s ∗ ⟦p⟧s = {< p >, < a >} ∗ p = {∅} Meaning in natural language: Paul is sleeping
Exercise 2: Predicate modification
Given are the following sets:
⟦female⟧s = {M arge, M aggie, Sharon, Lisa} ⟦male⟧s = {Kyle, Eric, Randy, Kenney, Sharon, Bart, Homer, Stan} ⟦child ⟧s = {Kyle, Eric, Bart, Stan, Lisa, M aggie, Kenney} ⟦adult⟧s = {Randy, Sharon, M arge, Homer} ⟦from⟧s = {< x, y > S x is f rom y} = {< Kyle, Southpark >, < Eric, Southpark >, < Randy, Southpark >, < M arge, Springf ield >, < Bart, Springf ield >, < Stan, Southpark >, < Homer, Springf ield >, < Kenney, Southpark >, < M aggie, Springf ield >, < Lisa, Springf ield >, < Sharon, Southpark >}
Compute
a. ⟦the [female [adult from Southpark]] ⟧s b. ⟦the [[female adult] [from Southpark]] ⟧s
and explain the similarity or difference between the results.
Solution:
The results will be identical since set intersection is associative: A ∩ (B ∩ C) = (A ∩ B) ∩ C
Exercise 3: Determiners and Quantifiers
Calculate the truth conditions for the sentence
⟦no student reads the book ⟧s
according to the definitions in chapter 6.1 of the book.
Solution:
⟦no student reads the book ⟧s = 1 iff ⟦reads the book ⟧s ∈ ⟦no student⟧s iff ⟦reads the book ⟧s ∈ (⟦no⟧s←Ð∗ ⟦student⟧s) iff ⟦reads the book ⟧s ∈ ({< X, Y > S X ∩ Y = ∅}←Ð∗ {x S x is a student in s}) iff ⟦reads the book ⟧s ∈ {Y S {x S x is a student in s} ∩ Y = ∅} iff (⟦reads⟧sÐ→∗ ⟦the book ⟧s) ∈ {Y S {x S x is a student in s} ∩ Y = ∅} iff (⟦reads⟧sÐ→∗ (⟦the⟧s(⟦book ⟧s))) ∈ {Y S {x S x is a student in s} ∩ Y = ∅} iff (⟦reads⟧sÐ→∗ (⟦the⟧s({x S x is a book in s}))) ∈ {Y S {x S x is a student in s} ∩ Y = ∅} iff (⟦reads⟧sÐ→∗ ({< X, y > S X = {y}}({x S x is a book in s}))) ∈ {Y S {x S x is a student in s} ∩ Y = ∅} iff (⟦reads⟧sÐ→∗ thebook) ∈ {Y S {x S x is a student in s} ∩ Y = ∅} iff ({< x, y > S x reads y in s}Ð→∗ thebook) ∈ {Y S {x S x is a student in s} ∩ Y = ∅} iff {x S x reads the book in s} ∈ {Y S {x S x is a student in s} ∩ Y = ∅} iff {x S x is a student in s} ∩ {x S x reads the book in s} = ∅
Intuitively, reading (1) should be false and reading (2) should be true. The task is to formally proof this intuition by calculation of truth conditions.
The model
Modified model M =< D, I >:
D = {A, B, C, D, a, b, c, d} I(Bild) = {a, b, c, d} I(W and) = {A, B, C, D} I(zierte) = {< a, A >, < b, B >, < c, C >< d, D >}
Note that this model now contains the additional picture d (included as the constant d in the domain D as well as in the interpretation of Bild ) decorating the wall D (included as the relation < d, D > in the interpretation of zierte).
In situ interpretation
will give us the linear reading.
⟦Ein Bild zierte jede Wand ⟧s = ⟦Ein Bild ⟧s ∗Q ⟦zierte jede Wand ⟧s
The first step is to higher-order-plug the QDP ⟦ein Bild ⟧s with the VP, meaning that the QDP is meant to fill out the subject position of the VP. We do not need an arrow here since the VP in which the object is already included leaves only one position open (namely the subject position: ⟦zierte jede Wand ⟧s = {x ∶ x decorated every wall in s}) so we don’t have to specify from what side we are plugging. The Q is the indicator for plugging with a set of sets. Note that at this point, it is not obligatory yet to apply higher-order plugging: We could as well apply the default plugging operation, meaning that the VP has to be an element of the QDP. This does, however, not work with objects that are QDPs, here we do have to make use of higher-order plugging. But it does not harm to also do so for the subject, if we do so, the sentence after plugging is
= {<> ∶ {x ∶ x ∈ ⟦zierte jede Wand ⟧s} ∈ ⟦ein Bild ⟧s}
which is the set of all 0-tuples such that the set of things that decorated every wall is an element of ⟦ein Bild ⟧s, which is equal to the stamement
= 1 iff ⟦zierte jede Wand ⟧s ∈ ⟦ein Bild ⟧s
meaning that the sentence turns out to be true iff the VP (a set of individuals, namely those that decorated every wall) is an element of the QDP (a set of sets, namely those predicates that apply to a picutre).
Next we have to further decompose the VP into the verb itself and the object. Since ⟦jede Wand ⟧s is a QDP as well, we have to apply
= 1 iff (⟦zierte⟧sÐ∗→Q⟦jede Wand ⟧s) ∈ ⟦ein Bild ⟧s
higher-order plugging between the verb ⟦zierte⟧s and the QDP object ⟦jede Wand ⟧s. Here we do need the arrow: The relation ⟦zierte⟧s now stands alone and has two argument positions open (⟦zierte⟧s = {< x, y > ∶ x decorated y in s}) of which ⟦jede Wand ⟧s is supposed to fill the object position; therefore, we need to indicate that we plug at the right side of the relation (y). What we get now is the following:
= 1 iff {x ∶ {y ∶ < x, y >∈ ⟦zierte⟧s} ∈ ⟦jede Wand ⟧s} ∈ ⟦ein Bild ⟧s = 1 iff {x ∶ {y ∶ x decorated y in s} ∈ ⟦jede Wand ⟧s} ∈ ⟦ein Bild ⟧s
What we have here is the set of all subjects for which it holds that the set of all objects which are in the decorating-relation with the subject must be an element of ⟦jede Wand ⟧s, and this set of subjects that are in the decorating-relation with every wall must again be an element of ⟦ein Bild ⟧s.
Now we have to check what is in contained in this set. We have to go through every possible value for x (for all decorators, i.e. the pictures), check what are in the inner set the values for y (the decorated, i.e. the walls) and whether they are in ⟦jede Wand ⟧s, and for those y that fulfil the condition we can put the corresponding x into our outer set, and this set must again be an element of ⟦ein Bild ⟧s. So let’s assign values vor x (the decorators) and then check what it is that they decorate:
For x = a ∶ ⟦a zierte jede Wand ⟧s = 1 iff {y ∶ a decorated y in s} ∈ ⟦every wall ⟧s = 1 iff {A} ∈ ⟦every wall ⟧s = false Therefore, a ∉ {x ∶ {y ∶ < x, y >∈ ⟦zierte⟧s ∈ ⟦jede Wand ⟧s}} Therefore, a ∉ {x ∶ {y ∶ x decorated y in s} ∈ ⟦jede Wand ⟧s} ∈ ⟦ein Bild ⟧s
Quantifier raising
will give us the inverted reading.
⟦Ein Bild zierte jede Wand ⟧s = ⟦jede Wand (^) y ⟧s ∗ ⟦Ein Bild zierte ty ⟧s
Moving the QDP out of the sentence results in set formation for the sentence it has been moved out, and the plugging operation turns out to be true iff the set of things that fulfil the VP of being decorated by a picture is an element of the QDP:
= 1 iff ⟦ein Bild zierte ty ⟧s ∈ ⟦jede Wand ⟧s = 1 iff {y ∶ ⟦ein Bild zierte y⟧s = 1 } ∈ ⟦jede Wand ⟧s
Now we again have to decompose the VP into the two-place relation and the subject:
= 1 iff {y ∶ {x ∶ < x, y >∈ ⟦zierte⟧s} ∈ ⟦ein Bild ⟧s} ∈ ⟦jede Wand ⟧s = 1 iff {y ∶ {x ∶ x decorated y in s} ∈ ⟦ein Bild ⟧s} ∈ ⟦jede Wand ⟧s
Here we have a very similar syntax to what we already got when applying higher-order plugging above, just the other way round: Here we have the set of objects such that the set of subjects that decorated it are an element of ⟦ein Bild ⟧s, and this set of objects that are decorated by a picture again has to be an element of ⟦every wall ⟧s.
Now we again have to assign values for y (the decorated, i.e. the walls) in the outer set, check what are in the inner set the x that decorated it (the decorators, i.e. the pictures), whether they fulfil the property of being in ⟦ein Bild ⟧s and if so, whether the set of walls that are decorated by a picture is in the semantics of ⟦jede Wand ⟧s. So let’s assign values for y:
For y = A: ⟦ein Bild ziert A⟧s = 1 iff {x ∶ x decorated A in s} ∈ ⟦ein Bild ⟧s = 1 iff {a} ∈ ⟦ein Bild ⟧s = 1 iff {a} ∩ {a, b, c, d} ≠ ∅ = true Therefore, A ∈ {y ∶ {x ∶ < x, y >∈ ⟦zierte⟧s} ∈ ⟦ein Bild ⟧s} ∈ ⟦jede Wand ⟧s Therefore, A ∈ {y ∶ {x ∶ x decorated y in s} ∈ ⟦ein Bild ⟧s}
So the wall A belongs to those walls that are decorated by at least one picture.
The same calculation holds for the other walls:
... For y = D: ⟦ein Bild ziert D⟧s = 1 iff {x ∶ x decorated D in s} ∈ ⟦ein Bild ⟧s = 1 iff {d} ∈ ⟦ein Bild ⟧s = 1 iff {d} ∩ {a, b, c, d} ≠ ∅ = true Therefore, D ∈ {y ∶ {x ∶ < x, y >∈ ⟦zierte⟧s} ∈ ⟦ein Bild ⟧s} ∈ ⟦jede Wand ⟧s Therefore, D ∈ {y ∶ {x ∶ x decorated y in s} ∈ ⟦ein Bild ⟧s}
Hence, the set of walls that are decorated by a picutre consists of all the walls A,B,C and D:
{y ∶ {x ∶ < x, y >∈ ⟦zierte⟧s} ∈ ⟦ein Bild ⟧s} ∈ ⟦jede Wand ⟧s = {y ∶ {x ∶ x decorated y in s} ∈ ⟦ein Bild ⟧s} = {A, B, C, D}
and this set has to be in ⟦every wall ⟧s, it must be an element of those predicates that apply to ev-
ery wall (and therefore the set of walls must be a subset of the elements being decorated by a picture):
= 1 iff {A, B, C, D} ∈ ⟦jede Wand ⟧s = 1 iff {A, B, C, D} ∈ {Y ∶ ⟦Wand ⟧s ⊆ Y } = 1 iff {A, B, C, D} ∈ {A, B, C, D} = true
We see that in fact, every wall fulfils the property of being decorated by a picture. Hene, we see that the inverted reading, in which it must hold that for every wall there must be at least one picture which decorated that wall, is true – which is again exactly what we wanted.
Alternative:
t
e
*
<>
(Ch{x : he bought x })
<<<>>>
P redM od
chocolate
<e>
the
<e>
eats
e
Klaus
Operator Op wird (wie bei QR) auf LF bewegt: Opi he bought ti, mit he bought ti vom Typ aus dem Satz wird (wie bei QR) die Menge gebildet: Ch{x: he bought x } vom Typ Op selbst (also Verk¨upfung zwischen NP und RS in Form von PredMod) fungiert als eine Art Identit¨atsfunktion vom Typ <> Semantik des PredMod-Operators: F¨ur alle P, Q vom Typ ist Op(P )(Q) (d.h. die Anwen- dung des Operators wird in zwei Teilschritte zergliedert und zuerst auf den RS angwendet, dann auf die NP!) diejenige Funktion F vom Typ , sodass f¨ur alle x vom Typ gilt: F (x) = 1 gdw P (x) und Q(x) = 1. erster Schirtt: PredMod-Operator vom Typ <<<>>> nimmt die Menge vom Typ als Argument und gibt als Wert die Identit¨atsfunktion vom Typ <> aus, die durch Funktionsanwendung auf chocolate wieder eine Eigenschaft macht zweiter Schritt: Anwendung der (den RS bereits enthaltenden) Identit¨atsfunktion vom Typ <> auf chocolate vom Typ ergibt die gePredModte NP vom Typ
Exercise 2: Composition of types
Explain the problem with compositional type assignment in the sentence
Klaus eats a chocolate
and one way this problem can be resolved.
Problem:
?? Type mismatch
<t>
chocolate
<<t>>
a
<e>
eats
e
Klaus
Solution 1:
In situ (type shifting ”eats”)
t
<t>
chocolate
<<t>>
a
<<t>>
<e>
eats
e
Klaus
Type-Shift des Verbs:
● Lif t(V f¨ur Indiv-Obj ) = V f¨ur QDP-Obj ● Lif t(<e>) = <<t>> ● zweist. Verb mit Indiv-Objekt als Argument (und Menge von Subjekten als Wert) wird angehoben zu zweist. Verb mit QDP-Objekt als Argument (und Menge von Subjekten als Wert)
Semantik des Lift-Operators:
● Altes Format (ohne Typen): Lif t(R)(QDP ) = {x ∶ {y ∶ < x, y > ∈ R} ∈ QDP } ● Neues Format (mit Typen): Lif t(R)(QDP ) = Ch{x ∶ QDP (Ch{y ∶ R(y)(x) = 1 }) = 1 } mit R vom Typ <e> und QDP vom Typ <t>
Eberhard Karls Universit¨at T¨ubingen | Seminar f¨ur Sprachwissenschaft (SfS) PS Semantik 1 (WS 2015/16), Prof. Feldgang Wolfstern Tutoren: Thorwald Kassemek, Natalie Clarius
Exercise Sheet 5:
Propositions and intensions
Zimmermann & Sternefeld: Ch. 7- Slides ESSLLI: No. 5
Exercise 1: Propositions
Consider the following worlds:
w 1 w 2 w 3 w 4
Determine
a. ⟦John loves Mary⟧
b. ⟦Mary loves John⟧
c. ⟦John loves Mary⟧w for each w
d. ⟦Mary loves John⟧w for each w
What are the propositions expressed by the sentences?
Solution:
a. ⟦John loves Mary⟧ = {w ∶ ⟦John loves Mary⟧w = 1 } = {w 1 , w 2 }
b. ⟦Mary loves John⟧ = {w ∶ ⟦Mary loves John⟧w = 1 } = {w 1 , w 3 }
c. ⟦John loves Mary⟧w 1 = 1 iff w 1 ∈ ⟦John loves Mary⟧ = 1
d. ⟦John loves Mary⟧w 2 = 1 iff w 2 ∈ ⟦John loves Mary⟧ = 1
e. ⟦John loves Mary⟧w 3 = 1 iff w 3 ∈ ⟦John loves Mary⟧ = 0
f. ⟦John loves Mary⟧w 4 = 1 iff w 4 ∈ ⟦John loves Mary⟧ = 0
g. ⟦Mary loves John⟧w 1 = 1 iff w 1 ∈ ⟦Mary loves John⟧ = 1
h. ⟦Mary loves John⟧w 2 = 1 iff w 2 ∈ ⟦Mary loves John⟧ = 0
i. ⟦Mary loves John⟧w 3 = 1 iff w 3 ∈ ⟦Mary loves John⟧ = 1
j. ⟦Mary loves John⟧w 4 = 1 iff w 4 ∈ ⟦Mary loves John⟧ = 0
The propositions of the sentences are the sets of posssible worlds of which that sentence is true.
