CSE541
Midterm 1
SOLUTIONS
Fall 2010
ˆL semantics
for ˆL
{¬
,
⇒
,
∩
,
∪}
is defined as follows
ˆL Negation
¬
F
⊥
T
T
⊥
F
ˆL Conjunction
∩
F
⊥
T
F
F
F
F
⊥
F
⊥
⊥
T
F
⊥
T
ˆL Disjunction
∪
F
⊥
T
F
F
⊥
T
⊥
⊥
⊥
T
T
T
T
T
ˆLImplication
⇒
F
⊥
T
F
T
T
T
⊥
⊥
T
T
T
F
⊥
T
QUESTION 1
(1)
Use the fact
that
v
:
V AR
→ {
F,
⊥
, T
}
be such that
v
*
((
a
∩
b
)
⇒ ¬
b
) =
⊥
under
ˆL
semantics
to evaluate
v
*
(((
b
⇒ ¬
a
)
⇒
(
a
⇒ ¬
b
))
∪
(
a
⇒
b
)).
Use shorthand notation.
(1) Solution
:
((
a
∩
b
)
⇒ ¬
b
) =
⊥
in two cases.
C1
(
a
∩
b
) =
⊥
and
¬
b
=
F
.
C2
(
a
∩
b
) =
T
and
¬
b
=
⊥
.
Case C1:
¬
b
=
F
, i.e.
b
=
T
, and hence (
a
∩
T
) =
⊥
iff
a
=
⊥
. We get that
v
is such that
v
(
a
) =
⊥
and
v
(
b
) =
T
.
We evaluate:
v
*
(((
b
⇒ ¬
a
)
⇒
(
a
⇒ ¬
b
))
∪
(
a
⇒
b
)) = (((
T
⇒ ¬ ⊥
)
⇒
(
⊥⇒
¬
T
))
∪
(
⊥⇒
T
)) = ((
⊥⇒⊥
)
∪
T
) =
T
.
Case C2:
¬
b
=
⊥
, i.e.
b
=
⊥
, and hence (
a
∩ ⊥
) =
T
what is impossible, hence
v
from case C1 is the only one.
(2)
Prove that in classical semantics
L
{¬
,
⇒}
≡ L
{¬
,
⇒
,
∪}
.
1
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We define
the EQUIVALENCE of LANGUAGES as follows:
Given two languages:
L
1
=
L
CON
1
and
L
2
=
L
CON
2
, for
CON
1
6
=
CON
2
.
We say that they are
logically equivalent
, i.e.
L
1
≡ L
2
if and only if the following conditions
C1, C2
hold.
C1:
For every formula
A
of
L
1
, there is a formula
B
of
L
2
, such that
A
≡
B,
C2:
For every formula
C
of
L
2
, there is a formula
D
of
L
1
, such that
C
≡
D.
(2) Solution:
(Classical case)
C1
holds because any formula of
L
{¬
,
⇒}
is a formula of
L
{¬
,
⇒
,
∪}
.
C2
holds due to the following definability of connectives equivalence
(
A
∪
B
)
≡
(
¬
A
⇒
B
)
.
(3)
Prove that the equivalence defining
∪
in classical logic does not hold under
ˆL
semantics, but nevertheless
L
{¬
,
⇒}
≡ L
{¬
,
⇒
,
∪}
.
Solution
(
A
∪
B
)
6≡
ˆL
(
¬
A
⇒
B
) Take
A
=
B
=
⊥
.
We get
⊥ ∪ ⊥
=
⊥
and
¬ ⊥⇒⊥
=
⊥⇒⊥
=
T
.
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 Spring '08
 Bachmair,L
 Computer Science, Logic, ¬b, ¬a, Completeness Theorem, ¬b ⇒ ¬a

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