Semantics 1, SoSe 2016 (Sailer)

Additional material for week 6

Quantifiers

Video introducing determiners into our logical language. (The video is based on the scenario of Romeo and Juliett.)

(copied from Wiki-ch2#Logical_determiners.2Fquantifiers)

Exercises

After having watched the video, work on the following tasks.

Task 1 Identify the determiners in the following sentence.

(a) Juliet talked to some stranger at the party.

(b) Every Capulet is an enemy to some Montague.

(c) Many people in Verona are not happy about the Capulet-Montague feud.

Check your solutions here:

(a) some

(b) every, some

(c) many

Task 2 Identify the formula that corresponds to the translation of the sentence.

Some Montague who was at the party fell in love with Juliet.

	∃x (montague1(x) : (at-party1(x) ∧ fall-in-love-with2(x,juliet)))
	∃x ((montague1(x) ∧ at-party1(x)) : fall-in-love-with2(x,juliet))
	∃x (montague1(x) : (at-party1(x) ∧ fall-in-love-with2(x,juliet))
	∃x ((montague1(x) ∧ fall-in-love-with2(x,juliet)) : at-party1(x))

Task 3 The sentence: Some Tybalt loved some Montague. is translated into the formula
∃ y (montague₁(y) : love₂(tybalt,y).

Mark all the cells in the table that stand for a true statement.

montague1(y) zwisch	love2(tybalt,y)zwisch
		Romeo
		Mercutio
		Juliet
		Tybalt
		Laurence
		Paris

Given this table, is the overall formula true or false? (Give a reason for your answer.)

Check your solutions here:

The formula is false, because there is no individual in our model for which both the restrictor and the scope are true.

Task 4 Variable assignment function
Start with the following variable assigment function g: g(u) = Romeo, g(v) = Juliet, g(w) = Romeo, g(x) = Laurence, g(y) = Mercutio, g(z) = Juliet

Provide the changed variable assignment function g[v/Paris].

Check your solutions here:

g[v/Paris](u) = g(u) = Romeo
g[v/Paris](v) = Paris
g[v/Paris](w) = g(w) = Romeo
g[v/Paris](x) = g(x) = Laurence
g[v/Paris](y) = g(y) = Mercutio
g[v/Paris](z) = g(z) = Juliet

Additional material for week 5

Formulae with more than one connective

The video shows how the truth value of a more complex formula can be computed. The example contains two connectives:

kill(malcom,lady-macbeth) ∨ ¬thane(macbeth)

The video shows two different methods: top down and bottom up.

Truth tables

(The following exercises have been copied here from the page on exercises for truth tables.)

Click on the boxes for which the truth value would be true.

p	q	zwisch(p ∧ q)zwisch	zwisch ¬(p ∧ q)zwisch	zwisch(q ⊃ ¬ (p ∧ q))zwisch
					(both p and q are true)
					(p is true, but not q)
					(p is false, but q is true)
					(both p and q are false)

Click on the boxes for which the truth value would be true.

p	q	r	zwisch¬ rzwisch	zwisch (p ∨ q)zwisch	zwisch((p ∨ q) ⊃ ¬r)zwisch
						(p, q, and r are true)
						(p and q are true, r is false)
						(p and r are true, q is false)
						(p is true, q and r are false)
						(p is false, q and r are true)
						(p and r are false, q is true)
						(p and q are false, r is true)
						(p, q, and r are false)

Truth tables for complex formulae

Truth tables are also useful to compute the truth value of complex formulae. This is shown in the following podcast, created by Lisa Günthner.

Preparation for week 5

• Read Levine et al (in prep.), Chapter 2, section 2.
• Using your model from last week,

• Give 1 formula with ⊃.
• Give 1 formule with 2 different connectives (both distinct from ⊃)

• Provide the step-by-step computation of the truth of your 2 
formulae.

Additional material for week 4

The material can be found on the page Semantics 1, SoSe 2016 (Sailer): Week 4

Additional material for week 3

The material for week 3 can be accessed here