20
Solutions to Exercises
p
0
pVO
8. The truth table shows that
p
y..
0
<=>
p:
T
F
T
F
F
F
p
,p
p
V
(,p)
1
9. [BBJ The truth table shows that (
P
Y
(,p»
<===?
1:
T
F
T
T
F
T
T
T
12. [BBJ This is no longer true:
Finally, neither law of absorption holds:
p
y..
(p/\q)
is not logically equivalent to
p andp/\ (p
y..
q)
is not logically equivalent
to
p.
I
~
I
; I
p; q
I
p
V
~
/\
q)
I I
~
I
; I
p; q
I
p
/\
~
V
q)
I
9. (a)
(p
V
q)
/\
«,p)
V
(,q))
is notin disjunctive normal form because the terms are not joined by V.
(b) [BBJ
(p
/\
q)
V
«,p)
/\
(,q»
is in disjunctive normal form.
(c) [BBJ
p
V
«,p)
/\
q)
is notin disjunctive normal form: not all variables are included in the first
term.
(d)
(p
/\
q)
V
«
,p)
/\
(,q) /\.r)
is not in disjunctive normal form: not all variables are included in the
first term.
.
(e)
(p
/\
q
/\
r)
V
«
,p)
/\
(,q)
/\
(,r))
is in disjunctive normal form.
10. (a) [BBJ This is already in disjunctive normal form! The definition permits just a single midterm.
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 Summer '10
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 Logic, Graph Theory, Conjunctive normal form, disjunctive normal form

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