PRACTICE FINAL
1.
True or False? (2 pts each)
L
is a firstorder language (with equality).
M
and
M
0
are interpreta
tions of
L
.
X
M
is the domain of
M
.
X
M
0
is the domain of
M
0
.
ψ
(
x
) is
a formula in
L
that has
x
as its only free variable.
θ
[
x
:
t
] is obtained
from
θ
by replacing every free occurrence of
x
by
t
.
•
If
X
M
has more elements than
X
M
0
and
X
M
is finite, then there
is a
φ
∈
L
such that
M

=
φ
and
M
0

=
¬
φ
.
True
:
As we discussed, for each natural number
k
there is a
sentence
λ
k
such that
M

=
λ
k
iff
X
M
has at least
k
elements.
For example, we can take
λ
2
to be
∃
x
∃
y
¬
(
x
≈
y
).
•
If
A
and
B
are
L
definable subsets of
X
M
, then
A
×
B
is an
L
definable subset of
X
M
×
X
M
.
True
: If
φ
(
x
) defines
A
and
ψ
(
y
) defines
B
, then
φ
(
x
)
∧
ψ
(
y
)
defines the relation
A
×
B
.
•
If
M

=
∀
xψ
(
x
), then
X
M
is the subset of
X
M
that is defined
by
ψ
(
x
).
True
: This follows directly from the definitions.
•
If
M

=
∃
xα
∧ ∃
xβ
, then
M

=
∃
x
(
α
∧
β
).
False
: Suppose
L
is built from a unary relation symbol
H
. Let
X
M
=
{
0
,
1
}
. Let
H
M
=
{
0
}
.
M

=
∃
xH
(
x
)
∧ ∃
x
¬
H
(
x
) but it
is not the case that
M

=
∃
x
(
H
(
x
)
∧ ¬
H
(
x
)).
•
If
a
and
b
are constant symbols in
L
and
M

=
θ
[
x
:
a
]
↔
θ
[
x
:
b
]
for all
θ
, where
θ
has
x
as its only free variable, then
M

=
a
≈
b
.
True
: Suppose that
M

=
θ
[
x
:
a
]
↔
θ
[
x
:
b
] for all
θ
, where
θ
has
x
as its only free variable. It follows that
M

= (
x
≈
a
)[
x
:
a
]
↔
(
x
≈
a
)[
x
:
b
]
but since
M

= (
x
≈
a
)[
x
:
a
] it follows that
M

= (
x
≈
a
)[
x
:
b
].
Thus,
M

=
a
≈
b
.
2.
Explain (5 pts each)
Explain two of your answers from Section 1.
1
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PRACTICE FINAL
3.
Inductive Definition (5 pts each)
Using the successor function, give an inductive definition of the set
{
(
q, r, s
)

q, r, s
∈
N
and
q
+
r
≤
s
}
.
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 Spring '11
 H
 Equivalence relation, Binary relation, Cartesian product, Finitary relation, unary relation symbol

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