Phil1001__Problem_Set_12___Answer_Key
pdf
keyboard_arrow_up
School
University of Minnesota-Twin Cities *
*We aren’t endorsed by this school
Course
1001
Subject
Philosophy
Date
Apr 3, 2024
Type
Pages
6
Uploaded by KidBraveryLemur17
Phil1001: Introduction to Logic
Name:
Fall 2021
Group:
Problem Set 12: First-Order Semantics - Solutions
Due: December 13, 2021 at 11:59pm
These are example answers, but yours may differ. Ask your TA if you are uncertain about the
quality of your answer.
For questions 1 and 2, we’ll consider these two predicates, with the following intended readings:
E
(
x, y
)
:
x
envies
y
B
(
x
)
:
x
is brilliant
And the following names:
s, m, w
1. (2pts)
Giving interpretations.
Given the domain
characters in the world of Sherlock Holmes
,
finish giving an interpretation of the predicates and names above. (Here’s a Wiki link to
Sherlock, if you’re unfamiliar!)
domain: characters in the world of Sherlock Holmes
s
:
Sherlock
w
:
Watson
m
:
Moriarty
‘
E
(
x, y
)
′
is true of
<
Watson, Sherlock
>, <
Moriarty, Sherlock
>, <
Sherlock, Moriarty
>
‘
B
(
x
)
’ is true of Sherlock and Moriarty
2. (2pts)
Giving interpretations.
Now, choose a different domain! (Unrelated to Sherlock.)
Using this new domain, give an interpretation of the same predicates and names. You do not
need to keep the intended readings (’x envies y’ and ’x is brilliant’), but you may if you wish.
domain: me and my siblings
s
:
Cat
w
:
Joey
m
:
Marty
‘
E
(
x, y
)
′
takes on the intended reading ‘x eats food off of y’s plate’ and is true of
<
Marty,
Joey
>, <
Joey, Marty
>, <
Marty, Marty
>, <
Joey, Joey
>, <
Cat, Cat
>, <
Marty, Cat
>, <
Joey,
Cat
>
‘
B
(
x
)
’ takes on the intended reading ‘x is a brother’ and is true of Marty and Joey.
For questions 3 and 4, consider the following predicates:
Y(x, y), S(x)
And the following names:
a, b, c, d
And the following interpretation:
domain: the Hemsworth brothers (Liam, Chris, and Luke)
a
: Liam Hemsworth
b
: Chris Hemsworth
c
: Luke Hemsworth
d
: Chris Hemsworth
‘Y(x, y)’ is true of
<
Liam, Chris
>, <
Chris, Luke
>, <
Liam, Luke
>
. (You can read this as
x
is younger than
y
)
‘S(x)’ is true of Liam, Chris, and Luke. (You can read this as ‘
x
has a sibling’.)
3. (3pts)
Semantics for FOL: Atomic & Sentential Sentences
For each of the following, indi-
cate (1) Whether it is a sentence of FOL, given the predicates and names we are considering,
and (2) whether it is true or false. Justify your answer to (2) using the semantics for introduced
in
§
30.1 and
§
30.2 of the text.
a.
S
(Liam)
(1) Is this a sentence of FOL?
No, because ‘Liam’ is not a term in FOL.
(2) If yes, then: Is it true? How do you know?
b.
S
(
a
)
(1) Is this a sentence of FOL?
Yes
(2) If yes, then: Is it true? How do you know?
Yes, because the name ‘a’ refers to Liam, and Liam is in the extension of ‘S’ on our
interpretation, which is the truth condition for atomic sentences of this form.
c.
b
=
d
(1) Is this a sentence of FOL?
Yes
(2) If yes, then: Is it true? How do you know?
Yes, because the names ‘b’ and ‘d’ both pick out Chris Hemsworth on the interpre-
tation, and ‘=’ sentences are true just when the names pick out the same object.
d.
(
S
(
S
(
b
))
∧
S
(
a
))
(1) Is this a sentence of FOL?
No, because predicates can only be applied to terms, but
(
S
(
b
))
∧
S
(
a
)
is not a term.
(2) If yes, then: Is it true? How do you know?
e.
(
S
(
a
)
∧
S
(
b
))
∧
Y
(
a, b
)
2
(1) Is this a sentence of FOL?
Yes
(2) If yes, then: Is it true? How do you know?
Yes, because the semantic entry for the main connective, conjunction, states that
the conjunction is true
iff
both conjuncts
(
S
(
a
)
∧
S
(
b
))
and
Y
(
a, b
)
are true on the
intepretation.
(
S
(
a
)
∧
S
(
b
))
is true on the interpretation
iff
S
(
a
)
and
S
(
b
)
are true on the inter-
pretation.
S
(
a
)
and
S
(
b
)
are true on the interpretation
iff
the objects named by ‘a’ and ‘b’
respectively are in the extension of S on the interpretation.
’a’ = Liam and ‘b’ = Chris, and both are in the extension of S.
So, following the chain of biconditionals up, we know that
(
S
(
a
)
∧
S
(
b
))
is true.
Looking at the right-hand side,
Y
(
a, b
)
is true
iff
the pair picked out by ‘a’ and
‘b’ is in the extension of Y on the interpretation.
’a’ = Liam and ‘b’ = Chris, and the pair
<
Liam, Chris
>
is in the extension of Y.
So, following the chain of biconditionals up, we know that
Y
(
a.b
)
is true.
Now, we have shown that both
(
S
(
a
)
∧
S
(
b
))
and
Y
(
a, b
)
are true, so we can
follow the chain of biconditionals up, and we have shown that the conjunction
(
S
(
a
)
∧
S
(
b
))
∧
Y
(
a, b
)
is true.
f.
(
Y
(
d, a
)
→
d
=
b
)
(1) Is this a sentence of FOL?
Yes.
(2) If yes, then: Is it true? How do you know?
Yes. To see this, consider the semantic entry for the conditional.
(
Y
(
d, a
)
→
d
=
b
)
is true
iff
either
Y
(
d, a
)
is false or
d
=
b
is true in the interpre-
tation.
Y
(
d, a
)
is true in the interpretation
iff
the pair picked out by ‘d’ and ‘a’, in that
order, is in the extension of Y in the interpretation.
‘d’ = Chris and ‘a’ = Liam, but
<
Chris, Liam
>
is not in the extension of Y in the
interpretation.
So,
Y
(
d, a
)
is false in the interpretation.
Following the chain of biconditionals up, we have shown that the interpretation
satisfies the truth condition for the conditional, so we have shown that
(
Y
(
d, a
)
→
d
=
b
)
is true on the interpretation.
3
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
4. (3pts)
Semantics for FOL: Quantified Sentences
Determine whether each of the follow-
ing quantified sentences is true on the above interpretation. Justify your answer using the
semantics for quantifiers introduced in
§
30.3 of the text.
1.
∀
xS
(
x
)
Yes. We can see that this is true by looking at the semantic entry for
∀
.
∀
xS
(
x
)
is true on an interpretation
I
iff
every object d in the domain satisfies
S
(
c
)
on
I
[
d/c
]
, for arbitrary c and d, where c is a name that is not already used in the interpre-
tation.
Since our domain has three objects, Liam, Chris, and Luke, we’ll check all three.
Suppose we have
I
[
Liam/l
]
, so that the name ‘l’ picks out Liam.
S
(
l
)
is true on
I
[
Liam/l
]
iff
the object picked out by ‘l’ is in the extension of S on
I
[
Liam/l
]
.
‘l’ picks out Liam and Liam is in the extension of S in
I
[
Liam/l
]
. So, following the bi-
conditional,
S
(
l
)
is true on
I
[
Liam/l
]
Suppose we have
I
[
Chris/r
]
, so that the name ‘r’ picks out Chris.
S
(
r
)
is true on
I
[
Chris/r
]
iff
the object picked out by ‘r’ is in the extension of S on
I
[
Chris/r
]
.
‘r’ picks out Chris and Chris is in the extension of S in
I
[
Chris/r
]
. So, following the
biconditional,
S
(
r
)
is true on
I
[
Chris/r
]
Suppose we have
I
[
Luke/u
]
, so that the name ‘u’ picks out Luke.
S
(
u
)
is true on
I
[
Luke/u
]
iff
the object picked out by ‘u’ is in the extension of S on
I
[
Like/u
]
.
‘u’ picks out Luke and Luke is in the extension of S in
I
[
Luke/u
]
. So, following the
biconditional,
S
(
u
)
is true on
I
[
Luke/u
]
We have now shown that every object d in the domain satisfies
S
(
c
)
on
I
[
d/c
]
.
Following the biconditionals, we have shown that
∀
xS
(
x
)
is true on this interpretation.
2.
∀
xY
(
x, x
)
No. We can see that this is false by looking at the semantic entry for
∀
.
∀
xY
(
x, x
)
is true on an interpretation
I
iff
every object d in the domain satisfies
Y
(
c, c
)
on
I
[
d/c
]
, for arbitrary c and d, where c is a name that is not already used in the inter-
pretation.
But, suppose we have an interpretation
I
[
Liam/m
]
.
Y
(
m, m
)
is true in
I
[
Liam/m
]
iff
the pair
< Liam, Liam >
is in the extension of Y
under
I
[
Liam/m
]
.
< Liam, Liam >
is not in the extension of Y under
I
[
Liam/m
]
So, following the biconditionals, we have shown that
∀
xY
(
x, x
)
is
not
true on this in-
terpretation.
3.
∃
yY
(
y, a
)
4
No. We can see that this is false by looking at the semantic entry for
∃
.
∃
yY
(
y, a
)
is true on an interpretation
I
iff
there is an object d in the domain such that
Y
(
k, a
)
is true on
I
[
d/k
]
, where k is a name that has not been used in
I
.
But,
Y
(
k, a
)
is true on
I
[
d/k
]
iff
the pair picked out by the objects named by ‘k’ and ‘a’
is in the extension of
Y
under
I
[
d/k
]
.
However, any such
I
[
d/k
]
will share the extension of Y under
I
, and there is no pair
in the extension if Y under
I
such that Liam, the object picked out by ‘a’, occupies the
second place in the ordered pair.
Therefore, there is no interpretation
I
[
d/k
]
that such that the pair picked out by the
objects named by ‘k’ and ‘a’ is in the extension of
Y
under
I
[
d/k
]
.
Following the biconditionals, then, we have shown that
∃
yY
(
y, a
)
is false under this
interpretation.
4.
∃
x
∀
yY
(
y, x
)
No. We can see that this is false by looking at the semantic entry for
∃
.
∃
x
∀
yY
(
y, x
)
is true on an interpretation
I
iff
there is an object d in the domain such
that
∀
yY
(
y, k
)
is true on
I
[
d/k
]
, where k is a name that has not been used in
I
.
Let
I
[
Luke/u
]
be such an interpretation.
∀
yY
(
y, u
)
is true on
I
[
Luke/u
]
iff
every object d in the domain satisfies
Y
(
g, u
)
on
I
[
Luke/u
][
d/g
]
, for arbitrary g and d, where g is a name that is not already used in the
interpretation.
But,
I
[
Luke/u
][
Luke/h
]
is one such interpretation.
And,
Y
(
h, u
)
is true on
I
[
Luke/u
][
Luke/h
]
iff
the pair picked out by the objects named
by ‘h’ and ‘u’, in that order, is in the extension of Y under
I
[
Luke/u
][
Luke/h
]
.
But, the pair
<
Luke, Luke
>
is not in the extension of Y under
I
[
Luke/u
][
Luke/h
]
, since
the extension of Y under
I
[
Luke/u
][
Luke/h
]
is the same as the extension of Y under
I
.
So, following the biconditionals,
∀
yY
(
y, u
)
is false on
I
[
Luke/u
]
.
And, we can carry out the same reasoning for each of the other objects in the domain:
there is no object d in the domain such that the pair
< d, d >
in the extension of Y under
I
. But, as we have just seen, there must be such a pair in order to make
∃
x
∀
yY
(
y, x
)
true.
5.
∃
x
∀
y
(
¬
x
=
y
→
Y
(
y, x
))
Yes. We can see this by examining the semantic entry for
∃
.
∃
x
∀
y
(
¬
x
=
y
→
Y
(
y, x
))
is true on an interpretation
I
iff
there is an object d in the
domain such that
∀
yY
(
y, k
)
is true on
I
[
d/k
]
, where k is a name that has not been used
in
I
.
Let
I
[
Luke/u
]
be such an interpretation.
∀
y
(
¬
u
=
y
→
Y
(
y, u
))
is true on
I
[
Luke/u
]
iff
every object d in the domain satisfies
¬
u
=
g
→
Y
(
g, u
)
on
I
[
Luke/u
][
d/g
]
, for arbitrary g and d, where g is a name that is
not already used in the interpretation.
Now, since there are three objects in the domain, we must consider three possible in-
5
terpretations:
I
[
Luke/u
][
Luke/h
]
I
[
Luke/u
][
Liam/h
]
I
[
Luke/u
][
Chris/h
]
First, we consider
I
[
Luke/u
][
Luke/h
]
.
¬
u
=
h
→
Y
(
h, u
)
is true on
I
[
Luke/u
][
Luke/h
]
iff
either
¬
u
=
h
is false or
Y
(
h, u
)
is true on
I
[
Luke/u
][
Luke/h
]
.
¬
u
=
h
is true in
I
[
Luke/u
][
Luke/h
]
iff
u
=
h
is false in
I
[
Luke/u
][
Luke/h
]
.
u
=
h
is true in
I
[
Luke/u
][
Luke/h
]
iff
the object picked out by ‘u’ and the object
picked out by ‘h’ are the same in
I
[
Luke/u
][
Luke/h
]
.
‘u’ and ‘h’ both pick out Luke in
I
[
Luke/u
][
Luke/h
]
.
Following the biconditionals, then,
¬
u
=
g
→
Y
(
g, u
)
is true on
I
[
Luke/u
][
Luke/h
]
.
Next, we consider
I
[
Luke/u
][
Liam/h
]
.
¬
u
=
h
→
Y
(
h, u
)
is true on
I
[
Luke/u
][
Liam/h
]
iff
either
¬
u
=
h
is false or
Y
(
h, u
)
is true on
I
[
Luke/u
][
Liam/h
]
.
Y
(
h, u
)
is true on
I
[
Luke/u
][
Liam/h
]
iff
< Liam, Luke >
is in the extension of Y
under
I
[
Luke/u
][
Liam/h
]
.
< Liam, Luke >
is in the extension of Y under
I
[
Luke/u
][
Liam/h
]
.
So, following the biconditionals,
¬
u
=
h
→
Y
(
h, u
)
is true on
I
[
Luke/u
][
Liam/h
]
.
Finally, we can follow the same reasoning for
I
[
Luke/u
][
Chris/h
]
,
mutatis mutandis
.
So, we have shown that
∃
x
∀
y
(
¬
x
=
y
→
Y
(
y, x
))
is true on this interpretation.
6.
∃
xY
(
x, x
)
∨ ¬∃
x
∀
yY
(
y, x
)
∃
xY
(
x, x
)
∨¬∃
x
∀
yY
(
y, x
)
is true on an interpretation
I
iff
either
∃
xY
(
x, x
)
or
¬∃
x
∀
yY
(
y, x
)
is true on I.
¬∃
x
∀
yY
(
y, x
)
is true on
I
iff
∃
x
∀
yY
(
y, x
)
is false in
I
.
In problem 4.4, we showed that
∃
x
∀
yY
(
y, x
)
is false on
I
.
Following the biconditionals, then, we have shown that
∃
xY
(
x, x
)
∨ ¬∃
x
∀
yY
(
y, x
)
is
true on this interpretation.
6
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help