MODULE
e
ONE
e
PROBLEM
e
SET
1
Directions:
e
Type
e
your
e
solutions
e
into
e
this
e
document
e
and
e
be
e
sure
e
to
e
show
e
all
e
steps
e
for
e
arriving
e
at
e
your
e
solution.
e
Just
e
giving
e
a
e
final
e
number
e
may
e
not
e
receive
e
full
e
credit.
P
ROBLEM
e
1
In
e
the
e
following
e
question,
e
the
e
domain
e
of
e
discourse
e
is
e
a
e
set
e
of
e
male
e
patients
e
in
e
a
e
clinical
e
study.
e
Define
e
the
e
following
e
predicates:
•
Pe (x)e :e xe wase givene thee placebo
•
D(x)
e
:
e
x
e
was
e
given
e
the
e
medication
•
M
e
(x)
e
:
e
x
e
had
e
migraines
Translate
e
each
e
of
e
the
e
following
e
statements
i
e
into
e
a
e
logical
e
expression.
e
Then
e
negate
e
the
e
expression
e
by
e
adding
e
a
e
negation
e
operation
e
to
e
the
e
beginning
i
e
of
e
the
i
e
expression.
e
Apply
e
De
e
Morgan’s
e
law
e
until
e
each
e
negation
e
operation
e
applies
i
e
directly
e
to
i
e
a
e
predicate
e
and
e
then
e
translate
e
the
i
e
logical
e
expression
e
back
i
e
into
e
English.
Sample
e
question:
e
Some
e
patient
e
was
e
given
e
the
e
placebo
i
e
and
e
the
i
medication.
•
∃
x
e
(P
e
(x)
e
∧
e
D(x))
•
Negation:
e
¬∃
x
e
(P
e
(x)
e
∧
e
D(x))
•
Applying
e
De
e
Morgan’s
e
law:
e
A
x
e
(
¬
P
e
(x)
e
∨
e
¬
D(x))
•
English:
e
Every
e
patient
e
was
e
either
e
not
e
given
e
the
e
placebo
e
or
e
not
e
given
e
the
e
medication
e
(or
e
both).
(a)
Every
e
patient
e
was
e
given
e
the
e
medication
e
or
e
the
e
placebo
e
or
e
both.
A
e
x
e
(D(x)
e
∨
e
P
e
(x))
Negation:
¬A
x
e
(D(x)
e
∨
e
P
e
(x))
Applying
e
De
e
Morgan’s
e
law:
∃
e
x(
¬
e
D(x)
e
∧
e
¬
P
e
(x))
English:
e
There
i
e
is
e
a
e
patient
e
who
i
e
was
e
not
i
e
given
e
the
e
medication
e
and
e
not
giventhe
e
placebo.
(b)
Everye patiente whoe tooke thee placeboe hade migraines.e (Hint:e youe wille
neede toe
apply
e
the
e
conditional
e
identity,
e
p
e
→
e
q
e
≡
e
¬
p
e
∨
e
q.)
A
x
e
(P
e
(x)
e
−→
e
M
e
(x))
Negation:
¬A
x
e
(P
e
(x)
e
−→
e
M
e
(x))
Applying
e
Conditional
e
identity:
e
(P
e
(x)
e
Applying
e
Double
e
Negation;
e
¬¬
e
P
e
(x)
≡
e
Applying
e
De
e
Morgan’s
e
law:
e
∃
e
x
e
(P
e
(x)
−→
e
M
e
(x))
P
e
(x)
∧e ¬
Me (x))
≡
e
(
¬
P
e
(x)
∨
e
¬
M
e
(x))
English:
e
Some
e
patient
e
took
e
the
e
placebo
e
and
e
did
e
not
e
have
e
migraines.
(c)
There
e
is
e
a
e
patient
e
who
e
had
e
migraines
e
and
e
was
e
given
e
the
e
placebo.
∃
e
x
e
(M
e
(x)
e
∧
e
P
e
(x))
Negation:
¬∃
e
x
e
(M
e
(x)
e
∧
e
P
e
(x))
Applying
e
De
e
Morgan’s
e
law:
e
A
e
x
e
(
¬
e
M
e
(x)
e
∨
e
¬
e
P
e
(x))
English:
e
Every
e
patient
e
did
e
not
e
have
i
e e
migraines
e
or
e
did
e
not
e
take
e
the
i
e e
placebo.
P
ROBLEM
e
2
Use
i
e
De
e
Morgan’s
e
law
e
for
e
quantified
e
statements
e
and
e
the
i
e
laws
e
of
e
propositionallogic
e
to
e
show
e
the
e
following
e
equivalences:
(a)
e
¬A
x
e
(P
e
(x)
e
∧
e
¬
Q(x))
e
≡
e
∃
x
e
(
¬
P
e
(x)
e
∨
e
Q(x))
¬A
x
e
(P
e
(x)
e
∧
e
¬
Q(x))
Applying
e
De
e
Morgan’s
e
Law:
e
∃
x
e
(
¬
(P
e
(x)
e
∨
e
¬¬
Q(x))
e
Applying
e
Double
e
Negation:
e
∃
x
e
(
¬
P
e
(x)
e
∨
e
Q(x))
e
Thereforee ∃xe (¬Pe (x)e ∨e Q(x))e ≡e ∃xe (¬Pe (x)e ∨e Q(x))
(b)
e
¬A
x
e
(
¬
P
e
(x)
e
→
e
Q(x))
e
≡
e
∃
x
e
(
¬
P
e
(x)
e
∧
e
¬
Q(x))
¬A
x
e
(
¬
P
e
(x)
e
−→
e
Q(x))
Applying
e
Conditional
e
identity:
e
(
¬¬
P
e
(x)
e
∨
e
Q(x))
e
Applying
e
Double
e
Negation:
e e
(P
e
(x)
e
∨
e e e
Q(x))
e
Applyinge Dee Morgan’se Law:e ∃xe (¬Pe (x)e ∨e ¬Q(x))
Thereforee ∃xe (¬Pe (x)e ∨e ¬Q(x))e ≡e ∃xe (¬Pe (x)e ∨e ¬Q(x))
(c)
e e
¬∃
x
e e
¬
P
e
(x)
e
∨
e
(Q(x)
e
∧
e
¬
R(x))
≡
e
A
x
e
P
e
(x)
e
∧
e
(
¬
Q(x)
e
∨
e
R(x))
¬∃xe (¬Pe (x)e ∨e (Q(x)e ∧e ¬R(x)))
Applying
e
De
e
Morgan’s
e
Law:
e
A
x
e
(P
e
(x)
e e
∧
e
¬
(Q(x)
e e
∧
e
¬
R(x)))
A
x
e
(P
e
(x)
e
∧
e
(
¬
Q(x)
e e
∨
e
¬¬
R(x)))
Applying
e
Double
e
Negation:
e
A
x
e
(P
e
(x)
e e
∧
e
(
¬
Q(x)
e e
∨
e
R(x)))
Therefore
e
A
x
e
(P
e
(x)
e
∧
e
(
¬
Q(x)
e e
∨
e
R(x)))
e e
≡
e e
A
x
e
(P
e
(x)
e e
∧
e
(
¬
Q(x)
e e
∨
e
R(x)))
P
ROBLEM
e
3
The
e
domain
e
of
e
discourse
e
for
e
this
i
e
problem
e
is
e
a
e
group
i
e
of
e
three
e
people
e
who
e
are
e
working
e
on
e
a
e
project.
e e
To
e
make
e
notation
e
easier,
e
the
i
e
people
e
are
i
e
numbered
e
1,
e
2,
e
3.
e
The
e
predicate
e
M
e
(x,
e e
y)
e
indicates
i
e
whether
e
x
i
e
has
i
e
sent
e
an
e
email
e
to
i
e
y,
e e
so
e
M
e
(2,
e e
3)
i
e
is
e
read
e
“Person
e
2
e
has
e
sent
i
e
an
e
email
e
to
e
person
e
3.”
e
The
e
table
i
e
below
e
shows
e
the
e
value
e
of
e
the
e
predicate
e
M
e
(x,
e
y)
i
e
for
e e
each
e e
(x,
e e
y)
e e
pair.
e e
The
e e
truth
e e
value
e e
in
e e
row
e e
x
e e
and
e
columny
e
gives
i
e
the
e
truth
e
value
i
e
for
e
M(x,
e
y).
M
1
2
3
1
T
T
T
2
T
F
T
3
T
T
F