SolutionsGroup5
On this page, you can find the solutions to the excercises on Quantifiers.
Restricted Quantifiers 1a
Sorry, this is not correct.
In restricted quantifier notation, the "complete" semantic representation of the noun phrase (NP) appears in the restrictor (-> square brackets).
Types Quantifiers 2d
Ramon signs every sculpture he makes.
2b) Universal Quantifier
∀x (SCULPTURE (x) & MAKE (r, x) → SIGN (r, x))
"For every thing x, if x is a sculpture and x is made by Ramon (r) then x is signed by Ramon (r)."
Since Ramon makes and signs the sculptures, the corresponding variables are (r, x).
2c) Restricted Quantifier
[Every x: SCULPTURE (x) & MAKE (r, x)] SIGN (r, x)
Here, the N' is "sculpture he makes" and therefore belongs in square brackets together with the Quantifier every.
Scopal Ambiguity 3a
In this sentence, the scopal ambiguity is created by the two quantifiers everyone and someone.
When looking at the two pictures that try to help you, you can see two possible readings:
1. For every person there is, there is at least one other person who loves him / her.
2. There is one person that is loved by everyone else.
Restricted Quantifiers 1c
Sorry, this is not correct.
In restricted quantifier notation, the semantic representation of the noun phrase (NP) appears in the restrictor.
Types Quantifiers 2a
Sorry, this is not correct.
Existential quantifiers are used for sentences that represent something that exists.
Of course, you could argue that there is a Person x such that x is called Ramon and x makes (and then signs) sculptures - but this is not what we were going for.
Maybe you want to check the possible answers once more.
Types Quantifiers 2b
Yes, this is correct.
Check if there is another correct answer!
Restricted Quantifiers 1b
Yes, this is correct.
Types Quantifiers 2c
Yes, this is correct.
Check if there is another correct answer!
Restricted Quantifiers 1d
Sorry, this is not correct.
In restricted quantifier notation, the semantic representation of the noun phrase (NP) appears in the restrictor.
Scopal Ambiguity 3b
1. For every person there is at least one person who loves him / her:
∀x (PERSON (x) → ∃y (PERSON (y) & LOVE (x, y))
2. There is one person that is loved by everyone:
∃x (PERSON (x) → ∀y (PERSON (y) & LOVE (y, x))
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