LIVRO_How to Prove It A Structured Approach by Daniel J Velleman

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fa
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In
tr
od
uc
ti
on
1
1
Se
nt
en
ti
al
L
og
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8
1.
1
D
ed
uc
tiv
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R
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es
8
1.
2
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ru
th
Ta
bl
es
14
1.
3
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ar
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bl
es
an
d
Se
ts
26
1.
4
O
pe
ra
tio
ns
on
Se
ts
34
1.
5
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he
C
on
di
tio
na
la
nd
B
ic
on
di
tio
na
lC
on
ne
ct
iv
es
43
2
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ua
nt
ifi
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ti
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L
og
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55
2.
1
Q
ua
nt
ifi
er
s
55
2.
2
E
qu
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al
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ce
s
In
vo
lv
in
g
Q
ua
nt
ifi
er
s
64
2.
3
M
or
e
O
pe
ra
tio
ns
on
Se
ts
73
3
P
ro
of
s
84
3.
1
Pr
oo
f
St
ra
te
gi
es
84
3.
2
Pr
oo
fs
In
vo
lv
in
g
N
eg
at
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d
C
on
di
tio
na
ls
95
3.
3
Pr
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fs
In
vo
lv
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Q
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s
10
8
3.
4
Pr
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In
vo
lv
in
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d
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ic
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ls
12
4
3.
5
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fs
In
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lv
in
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D
is
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ns
13
6
3.
6
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fs
14
6
3.
7
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E
xa
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Pr
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15
5
4
R
el
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16
3
4.
1
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Pa
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s
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C
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Pr
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ts
16
3
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2
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el
at
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17
1
4.
3
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bo
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18
0
4.
4
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el
at
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ns
18
9
vi
i
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vi
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4.
5
C
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20
2
4.
6
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ce
R
el
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21
3
5
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un
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22
6
5.
1
Fu
nc
tio
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22
6
5.
2
O
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-t
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on
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O
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o
23
6
5.
3
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ve
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of
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tio
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24
5
5.
4
Im
ag
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In
ve
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:A
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Pr
oj
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t
25
5
6
M
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m
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du
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26
0
6.
1
Pr
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f
by
M
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m
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n
26
0
6.
2
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E
xa
m
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26
7
6.
3
R
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27
9
6.
4
St
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In
du
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n
28
8
6.
5
C
lo
su
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s
A
ga
in
30
0
7
In
fin
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Se
ts
30
6
7.
1
E
qu
in
um
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ou
s
Se
ts
30
6
7.
2
C
ou
nt
ab
le
an
d
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nc
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nt
ab
le
Se
ts
31
5
7.
3
T
he
C
an
to
r–
Sc
hr
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–B
er
ns
te
in
T
he
or
em
32
2
A
pp
en
di
x
1:
So
lu
ti
on
s
to
Se
le
ct
ed
E
xe
rc
is
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32
9
A
pp
en
di
x
2:
P
ro
of
D
es
ig
ne
r
37
3
Su
gg
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ti
on
s
fo
r
F
ur
th
er
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ea
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ng
37
5
Su
m
m
ar
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of
P
ro
of
Te
ch
ni
qu
es
37
6
In
de
x
38
1
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Pr
ef
ac
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St
ud
en
ts
of
m
at
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m
at
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s
an
d
co
m
pu
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r
sc
ie
nc
e
of
te
n
ha
ve
tr
ou
bl
e
th
e
fir
st
tim
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th
ey
’r
e
as
ke
d
to
w
or
k
se
ri
ou
sl
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w
ith
m
at
he
m
at
ic
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pr
oo
fs
,b
ec
au
se
th
ey
do
n’
tk
no
w
th
e
“r
ul
es
of
th
e
ga
m
e.
”
W
ha
ti
s
ex
pe
ct
ed
of
yo
u
if
yo
u
ar
e
as
ke
d
to
pr
ov
e
so
m
et
hi
ng
?
W
ha
t
di
st
in
gu
is
he
s
a
co
rr
ec
t
pr
oo
f
fr
om
an
in
co
rr
ec
t
on
e?
T
hi
s
bo
ok
is
in
te
nd
ed
to
he
lp
st
ud
en
ts
le
ar
n
th
e
an
sw
er
s
to
th
es
e
qu
es
-
tio
ns
by
sp
el
lin
g
ou
tt
he
un
de
rl
yi
ng
pr
in
ci
pl
es
in
vo
lv
ed
in
th
e
co
ns
tr
uc
tio
n
of
pr
oo
fs
.
M
an
y
st
ud
en
ts
ge
t
th
ei
r
fir
st
ex
po
su
re
to
m
at
he
m
at
ic
al
pr
oo
fs
in
a
hi
gh
sc
ho
ol
co
ur
se
on
ge
om
et
ry
.
U
nf
or
tu
na
te
ly
,
st
ud
en
ts
in
hi
gh
sc
ho
ol
ge
om
et
ry
ar
e
us
ua
lly
ta
ug
ht
to
th
in
k
of
a
pr
oo
f
as
a
nu
m
be
re
d
lis
t
of
st
at
em
en
ts
an
d
re
as
on
s,
a
vi
ew
of
pr
oo
fs
th
at
is
to
o
re
st
ri
ct
iv
e
to
be
ve
ry
us
ef
ul
.
T
he
re
is
a
pa
ra
lle
lw
ith
co
m
pu
te
rs
ci
en
ce
he
re
th
at
ca
n
be
in
st
ru
ct
iv
e.
E
ar
ly
pr
og
ra
m
m
in
g
la
ng
ua
ge
s
en
co
ur
ag
ed
a
si
m
ila
rr
es
tr
ic
tiv
e
vi
ew
of
co
m
pu
te
rp
ro
gr
am
s
as
nu
m
-
be
re
d
lis
ts
of
in
st
ru
ct
io
ns
.
N
ow
co
m
pu
te
r
sc
ie
nt
is
ts
ha
ve
m
ov
ed
aw
ay
fr
om
su
ch
la
ng
ua
ge
s
an
d
te
ac
h
pr
og
ra
m
m
in
g
by
us
in
g
la
ng
ua
ge
s
th
at
en
co
ur
ag
e
an
ap
pr
oa
ch
ca
lle
d
“s
tr
uc
tu
re
d
pr
og
ra
m
m
in
g.
”
T
he
di
sc
us
si
on
of
pr
oo
fs
in
th
is
bo
ok
is
in
sp
ir
ed
by
th
e
be
lie
f
th
at
m
an
y
of
th
e
co
ns
id
er
at
io
ns
th
at
ha
ve
le
d
co
m
pu
te
r
sc
ie
nt
is
ts
to
em
br
ac
e
th
e
st
ru
ct
ur
ed
ap
pr
oa
ch
to
pr
og
ra
m
m
in
g
ap
-
pl
y
to
pr
oo
f-
w
ri
tin
g
as
w
el
l.
Y
ou
m
ig
ht
sa
y
th
at
th
is
bo
ok
te
ac
he
s
“s
tr
uc
tu
re
d
pr
ov
in
g.
”
In
st
ru
ct
ur
ed
pr
og
ra
m
m
in
g,
a
co
m
pu
te
rp
ro
gr
am
is
co
ns
tr
uc
te
d,
no
tb
y
lis
tin
g
in
st
ru
ct
io
ns
on
e
af
te
r
an
ot
he
r,
bu
t
by
co
m
bi
ni
ng
ce
rt
ai
n
ba
si
c
st
ru
ct
ur
es
su
ch
as
th
e
if
-e
ls
e
co
ns
tr
uc
ta
nd
do
-w
hi
le
lo
op
of
th
e
Ja
va
pr
og
ra
m
m
in
g
la
ng
ua
ge.
T
he
se
st
ru
ct
ur
es
ar
e
co
m
bi
ne
d,
no
to
nl
y
by
lis
tin
g
th
em
on
e
af
te
r
an
ot
he
r,
bu
t
al
so
by
ne
st
in
g
on
e
w
ith
in
an
ot
he
r.
Fo
r
ex
am
pl
e,
a
pr
og
ra
m
co
ns
tr
uc
te
d
by
ix
05
21
86
12
41
pr
e
C
B
99
6/
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m
an
O
ct
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P
re
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ith
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lte
rn
at
e
lis
to
f
in
st
ru
ct
io
ns
go
es
he
re
.]
w
hi
le
[c
on
di
tio
n]
T
he
in
de
nt
in
g
in
th
is
pr
og
ra
m
ou
tli
ne
is
no
t
ab
so
lu
te
ly
ne
ce
ss
ar
y,
bu
t
it
is
a
co
nv
en
ie
nt
m
et
ho
d
of
te
n
us
ed
in
co
m
pu
te
r
sc
ie
nc
e
to
di
sp
la
y
th
e
un
de
rl
yi
ng
st
ru
ct
ur
e
of
a
pr
og
ra
m
.
M
at
he
m
at
ic
al
pr
oo
fs
ar
e
al
so
co
ns
tr
uc
te
d
by
co
m
bi
ni
ng
ce
rt
ai
n
ba
si
c
pr
oo
f
st
ru
ct
ur
es
.F
or
ex
am
pl
e,
a
pr
oo
fo
fa
st
at
em
en
to
ft
he
fo
rm
“i
f
P
th
en
Q
”
of
te
n
us
es
w
ha
tm
ig
ht
be
ca
lle
d
th
e
“s
up
po
se
-u
nt
il”
st
ru
ct
ur
e:
W
e
su
pp
os
e
th
at
P
is
tr
ue
un
ti
lw
e
ar
e
ab
le
to
re
ac
h
th
e
co
nc
lu
si
on
th
at
Q
is
tr
ue
,a
tw
hi
ch
po
in
tw
e
re
tr
ac
tt
hi
s
su
pp
os
iti
on
an
d
co
nc
lu
de
th
at
th
e
st
at
em
en
t“
if
P
th
en
Q
”
is
tr
ue
.
A
no
th
er
ex
am
pl
e
is
th
e
“f
or
ar
bi
tr
ar
y
x
pr
ov
e”
st
ru
ct
ur
e:
To
pr
ov
e
a
st
at
em
en
t
of
th
e
fo
rm
“f
or
al
l
x,
P
(x
),
”
w
e
de
cl
ar
e
x
to
be
an
ar
bi
tr
ar
y
ob
je
ct
an
d
th
en
pr
ov
e
P
(x
).
O
nc
e
w
e
re
ac
h
th
e
co
nc
lu
si
on
th
at
P
(x
)
is
tr
ue
w
e
re
tr
ac
t
th
e
de
cl
ar
at
io
n
of
x
as
ar
bi
tr
ar
y
an
d
co
nc
lu
de
th
at
th
e
st
at
em
en
t“
fo
r
al
l
x,
P
(x
)”
is
tr
ue
.
Fu
rt
he
rm
or
e,
to
pr
ov
e
m
or
e
co
m
pl
ex
st
at
em
en
ts
th
es
e
st
ru
ct
ur
es
ar
e
of
te
n
co
m
bi
ne
d,
no
to
nl
y
by
lis
tin
g
on
e
af
te
r
an
ot
he
r,
bu
ta
ls
o
by
ne
st
in
g
on
e
w
ith
in
an
ot
he
r.
Fo
re
xa
m
pl
e,
to
pr
ov
e
a
st
at
em
en
to
ft
he
fo
rm
“f
or
al
lx
,i
f
P
(x
)
th
en
Q
(x
)”
w
e
w
ou
ld
pr
ob
ab
ly
ne
st
a
“s
up
po
se
-u
nt
il”
st
ru
ct
ur
e
w
ith
in
a
“f
or
ar
bi
tr
ar
y
x
pr
ov
e”
st
ru
ct
ur
e,
ge
tti
ng
a
pr
oo
f
of
th
is
fo
rm
:
L
et
x
be
ar
bi
tr
ar
y.
Su
pp
os
e
P
(x
)
is
tr
ue
.
[P
ro
of
of
Q
(x
)
go
es
he
re
.]
T
hu
s,
if
P
(x
)
th
en
Q
(x
).
T
hu
s,
fo
r
al
lx
,i
f
P
(x
)
th
en
Q
(x
).
A
s
be
fo
re
,w
e
ha
ve
us
ed
in
de
nt
in
g
to
m
ak
e
th
e
un
de
rl
yi
ng
st
ru
ct
ur
e
of
th
e
pr
oo
f
cl
ea
r.
O
fc
ou
rs
e,
m
at
he
m
at
ic
ia
ns
do
n’
to
rd
in
ar
ily
w
ri
te
th
ei
rp
ro
of
si
n
th
is
in
de
nt
ed
fo
rm
.
O
ur
ai
m
in
th
is
bo
ok
is
to
te
ac
h
st
ud
en
ts
to
w
ri
te
pr
oo
fs
in
or
di
na
ry
E
ng
lis
h
pa
ra
gr
ap
hs
,j
us
t
as
m
at
he
m
at
ic
ia
ns
do
,a
nd
no
t
in
th
e
in
de
nt
ed
fo
rm
.
N
ev
er
th
el
es
s,
ou
ra
pp
ro
ac
h
is
ba
se
d
on
th
e
be
lie
ft
ha
ti
fs
tu
de
nt
s
ar
e
to
su
cc
ee
d
at
w
ri
tin
g
su
ch
pr
oo
fs
,t
he
y
m
us
tu
nd
er
st
an
d
th
e
un
de
rl
yi
ng
st
ru
ct
ur
e
th
at
pr
oo
fs
ha
ve
.T
he
y
m
us
tl
ea
rn
,f
or
ex
am
pl
e,
th
at
se
nt
en
ce
s
lik
e
“L
et
x
be
ar
bi
tr
ar
y”
an
d
“S
up
po
se
P
”
ar
e
no
ti
so
la
te
d
st
ep
s
in
pr
oo
fs
,b
ut
ar
e
us
ed
to
in
tr
od
uc
e
th
e
“f
or
ar
bi
tr
ar
y
x
pr
ov
e”
an
d
“s
up
po
se
-u
nt
il”
pr
oo
f
st
ru
ct
ur
es
.
It
is
no
t
un
co
m
m
on
fo
r
be
gi
nn
in
g
st
ud
en
ts
to
us
e
th
es
e
se
nt
en
ce
s
in
ap
pr
op
ri
at
el
y
in
ot
he
r
w
ay
s.
05
21
86
12
41
pr
e
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
1
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
P
re
fa
ce
xi
Su
ch
m
is
ta
ke
s
ar
e
an
al
og
ou
s
to
th
e
pr
og
ra
m
m
in
g
er
ro
ro
fu
si
ng
a
“d
o”
w
ith
no
m
at
ch
in
g
“w
hi
le
.”
N
ot
e
th
at
in
ou
re
xa
m
pl
es
,t
he
ch
oi
ce
of
pr
oo
fs
tr
uc
tu
re
is
gu
id
ed
by
th
e
lo
g-
ic
al
fo
rm
of
th
e
st
at
em
en
tb
ei
ng
pr
ov
en
.F
or
th
is
re
as
on
,t
he
bo
ok
be
gi
ns
w
ith
el
em
en
ta
ry
lo
gi
c
to
fa
m
ili
ar
iz
e
st
ud
en
ts
w
ith
th
e
va
ri
ou
s
fo
rm
s
th
at
m
at
he
m
at
i-
ca
ls
ta
te
m
en
ts
ta
ke
.C
ha
pt
er
1
di
sc
us
se
s
lo
gi
ca
lc
on
ne
ct
iv
es
,a
nd
qu
an
tifi
er
s
ar
e
in
tr
od
uc
ed
in
C
ha
pt
er
2.
T
he
se
ch
ap
te
rs
al
so
pr
es
en
t
th
e
ba
si
cs
of
se
t
th
eo
ry
,
be
ca
us
e
it
is
an
im
po
rt
an
t
su
bj
ec
t
th
at
is
us
ed
in
th
e
re
st
of
th
e
bo
ok
(a
nd
th
ro
ug
ho
ut
m
at
he
m
at
ic
s)
,a
nd
al
so
be
ca
us
e
it
se
rv
es
to
ill
us
tr
at
e
m
an
y
of
th
e
po
in
ts
of
lo
gi
c
di
sc
us
se
d
in
th
es
e
ch
ap
te
rs
.
C
ha
pt
er
3
co
ve
rs
st
ru
ct
ur
ed
pr
ov
in
g
te
ch
ni
qu
es
in
a
sy
st
em
at
ic
w
ay
,r
un
ni
ng
th
ro
ug
h
th
e
va
ri
ou
sf
or
m
st
ha
tm
at
he
m
at
ic
al
st
at
em
en
ts
ca
n
ta
ke
an
d
di
sc
us
si
ng
th
e
pr
oo
f
st
ru
ct
ur
es
ap
pr
op
ri
at
e
fo
r
ea
ch
fo
rm
.T
he
ex
am
pl
es
of
pr
oo
fs
in
th
is
ch
ap
te
ra
re
fo
rt
he
m
os
tp
ar
tc
ho
se
n,
no
tf
or
th
ei
rm
at
he
m
at
ic
al
co
nt
en
t,
bu
tf
or
th
e
pr
oo
f
st
ru
ct
ur
es
th
ey
ill
us
tr
at
e.
T
hi
s
is
es
pe
ci
al
ly
tr
ue
ea
rl
y
in
th
e
ch
ap
te
r,
w
he
n
on
ly
a
fe
w
pr
oo
ft
ec
hn
iq
ue
s
ha
ve
be
en
di
sc
us
se
d,
an
d
as
a
re
su
lt
m
an
y
of
th
e
pr
oo
fs
in
th
is
pa
rt
of
th
e
ch
ap
te
ra
re
ra
th
er
tr
iv
ia
l.
A
s
th
e
ch
ap
te
rp
ro
gr
es
se
s
th
e
pr
oo
fs
ge
tm
or
e
so
ph
is
tic
at
ed
an
d
m
or
e
in
te
re
st
in
g,
m
at
he
m
at
ic
al
ly
.
C
ha
pt
er
s
4
an
d
5,
on
re
la
tio
ns
an
d
fu
nc
tio
ns
,
se
rv
e
tw
o
pu
rp
os
es
.
Fi
rs
t,
th
ey
pr
ov
id
e
su
bj
ec
t
m
at
te
r
on
w
hi
ch
st
ud
en
ts
ca
n
pr
ac
tic
e
th
e
pr
oo
f-
w
ri
tin
g
te
ch
ni
qu
es
fr
om
C
ha
pt
er
3.
A
nd
se
co
nd
,t
he
y
in
tr
od
uc
e
st
ud
en
ts
to
so
m
e
fu
n-
da
m
en
ta
lc
on
ce
pt
s
us
ed
in
al
lb
ra
nc
he
s
of
m
at
he
m
at
ic
s.
C
ha
pt
er
6
is
de
vo
te
d
to
a
m
et
ho
d
of
pr
oo
f
th
at
is
ve
ry
im
po
rt
an
t
in
bo
th
m
at
he
m
at
ic
s
an
d
co
m
pu
te
r
sc
ie
nc
e:
m
at
he
m
at
ic
al
in
du
ct
io
n.
T
he
pr
es
en
ta
tio
n
bu
ild
s
on
th
e
te
ch
ni
qu
es
fr
om
C
ha
pt
er
3,
w
hi
ch
st
ud
en
ts
sh
ou
ld
ha
ve
m
as
te
re
d
by
th
is
po
in
ti
n
th
e
bo
ok
.
Fi
na
lly
,i
n
C
ha
pt
er
7
m
an
y
id
ea
s
fr
om
th
ro
ug
ho
ut
th
e
re
st
of
th
e
bo
ok
ar
e
br
ou
gh
tt
og
et
he
r
to
pr
ov
e
so
m
e
of
th
e
m
os
td
if
fic
ul
ta
nd
m
os
ti
nt
er
es
tin
g
th
e-
or
em
s
in
th
e
bo
ok
.
I
w
ou
ld
lik
e
to
th
an
k
al
lt
ho
se
w
ho
re
ad
ea
rl
ie
r
dr
af
ts
of
th
e
m
an
us
cr
ip
ta
nd
m
ad
e
m
an
y
he
lp
fu
ls
ug
ge
st
io
ns
fo
ri
m
pr
ov
em
en
ts
,i
n
pa
rt
ic
ul
ar
L
au
re
n
C
ow
le
s
at
C
am
br
id
ge
U
ni
ve
rs
ity
Pr
es
s,
m
y
co
lle
ag
ue
Pr
of
es
so
r
D
ua
ne
B
ai
le
y
an
d
hi
s
D
is
cr
et
e
M
at
he
m
at
ic
s
cl
as
s,
w
ho
tr
ie
d
ou
t
ea
rl
ie
r
ve
rs
io
ns
of
so
m
e
ch
ap
te
rs
,
an
d
fin
al
ly
m
y
w
if
e,
Sh
el
le
y,
w
ith
ou
tw
ho
se
co
ns
ta
nt
en
co
ur
ag
em
en
tt
hi
s
bo
ok
w
ou
ld
ne
ve
r
ha
ve
be
en
w
ri
tte
n.
05
21
86
12
41
pr
e
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B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
1
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
xi
i
P
re
fa
ce
Pr
ef
ace
to
th
e
Se
co
nd
E
di
tio
n
I
w
ou
ld
lik
e
to
th
an
k
al
l
of
th
os
e
w
ho
ha
ve
se
nt
m
e
co
m
m
en
ts
ab
ou
t
th
e
fir
st
ed
iti
on
.T
ho
se
co
m
m
en
ts
ha
ve
re
su
lte
d
in
a
nu
m
be
ro
fs
m
al
lc
ha
ng
es
th
ro
ug
h-
ou
tt
he
te
xt
.H
ow
ev
er
,t
he
bi
gg
es
td
if
fe
re
nc
e
be
tw
ee
n
th
e
fir
st
ed
iti
on
an
d
th
e
se
co
nd
is
th
e
ad
di
tio
n
of
ov
er
20
0
ne
w
ex
er
ci
se
s.
T
he
re
is
al
so
an
ap
pe
nd
ix
co
nt
ai
ni
ng
so
lu
tio
ns
to
se
le
ct
ed
ex
er
ci
se
s.
E
xe
rc
is
es
fo
r
w
hi
ch
so
lu
tio
ns
ar
e
su
pp
lie
d
ar
e
m
ar
ke
d
w
ith
an
as
te
ri
sk
.I
n
m
os
tc
as
es
,t
he
so
lu
tio
n
su
pp
lie
d
is
a
co
m
pl
et
e
so
lu
tio
n;
in
so
m
e
ca
se
s,
it
is
a
sk
et
ch
of
a
so
lu
tio
n,
or
a
hi
nt
.
So
m
e
ex
er
ci
se
s
in
C
ha
pt
er
s
3
an
d
4
ar
e
al
so
m
ar
ke
d
w
ith
th
e
sy
m
bo
l
p d
.
T
hi
s
in
di
ca
te
s
th
at
th
es
e
ex
er
ci
se
s
ca
n
be
so
lv
ed
us
in
g
Pr
oo
f
D
es
ig
ne
r.
Pr
oo
f
D
es
ig
ne
r
is
co
m
pu
te
r
so
ft
w
ar
e
th
at
he
lp
s
th
e
us
er
w
ri
te
ou
tli
ne
s
of
pr
oo
fs
in
el
em
en
ta
ry
se
t
th
eo
ry
,
us
in
g
th
e
m
et
ho
ds
di
sc
us
se
d
in
th
is
bo
ok
.
Fu
rt
he
r
in
fo
rm
at
io
n
ab
ou
tP
ro
of
D
es
ig
ne
rc
an
be
fo
un
d
in
an
ap
pe
nd
ix
,a
nd
at
th
e
Pr
oo
f
D
es
ig
ne
r
w
eb
si
te
:h
t
t
p
:
/
/
w
w
w
.
c
s
.
a
m
h
e
r
s
t
.
e
d
u
/
∼d
j
v
/
p
d
/
p
d
.
h
t
m
l
.
05
21
86
12
41
in
t
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
18
,2
00
5
17
:2
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
In
tr
od
uc
tio
n
W
ha
ti
s
m
at
he
m
at
ic
s?
H
ig
h
sc
ho
ol
m
at
he
m
at
ic
s
is
co
nc
er
ne
d
m
os
tly
w
ith
so
lv
-
in
g
eq
ua
tio
ns
an
d
co
m
pu
tin
g
an
sw
er
s
to
nu
m
er
ic
al
qu
es
tio
ns
.C
ol
le
ge
m
at
he
-
m
at
ic
s
de
al
s
w
ith
a
w
id
er
va
ri
et
y
of
qu
es
tio
ns
,i
nv
ol
vi
ng
no
to
nl
y
nu
m
be
rs
,b
ut
al
so
se
ts
,f
un
ct
io
ns
,a
nd
ot
he
r
m
at
he
m
at
ic
al
ob
je
ct
s.
W
ha
t
tie
s
th
em
to
ge
th
er
is
th
e
us
e
of
de
du
ct
iv
e
re
as
on
in
g
to
fin
d
th
e
an
sw
er
s
to
qu
es
tio
ns
.W
he
n
yo
u
so
lv
e
an
eq
ua
tio
n
fo
r
x
yo
u
ar
e
us
in
g
th
e
in
fo
rm
at
io
n
gi
ve
n
by
th
e
eq
ua
tio
n
to
de
du
ce
w
ha
tt
he
va
lu
e
of
x
m
us
tb
e.
Si
m
ila
rl
y,
w
he
n
m
at
he
m
at
ic
ia
ns
so
lv
e
ot
he
r
ki
nd
s
of
m
at
he
m
at
ic
al
pr
ob
le
m
s,
th
ey
al
w
ay
s
ju
st
if
y
th
ei
r
co
nc
lu
si
on
s
w
ith
de
du
ct
iv
e
re
as
on
in
g.
D
ed
uc
tiv
e
re
as
on
in
g
in
m
at
he
m
at
ic
s
is
us
ua
lly
pr
es
en
te
d
in
th
e
fo
rm
of
a
pr
oo
f.
O
ne
of
th
e
m
ai
n
pu
rp
os
es
of
th
is
bo
ok
is
to
he
lp
yo
u
de
ve
lo
p
yo
ur
m
at
he
m
at
ic
al
re
as
on
in
g
ab
ili
ty
in
ge
ne
ra
l,
an
d
in
pa
rt
ic
ul
ar
yo
ur
ab
ili
ty
to
re
ad
an
d
w
ri
te
pr
oo
fs
.I
n
la
te
r
ch
ap
te
rs
w
e’
ll
st
ud
y
ho
w
pr
oo
fs
ar
e
co
ns
tr
uc
te
d
in
de
ta
il,
bu
tfi
rs
tl
et
’s
ta
ke
a
lo
ok
at
a
fe
w
ex
am
pl
es
of
pr
oo
fs
.
D
on
’t
w
or
ry
if
yo
u
ha
ve
tr
ou
bl
e
un
de
rs
ta
nd
in
g
th
es
e
pr
oo
fs
.
T
he
y’
re
ju
st
in
te
nd
ed
to
gi
ve
yo
u
a
ta
st
e
of
w
ha
t
m
at
he
m
at
ic
al
pr
oo
fs
ar
e
lik
e.
In
so
m
e
ca
se
s
yo
u
m
ay
be
ab
le
to
fo
llo
w
m
an
y
of
th
e
st
ep
s
of
th
e
pr
oo
f,
bu
ty
ou
m
ay
be
pu
zz
le
d
ab
ou
tw
hy
th
e
st
ep
s
ar
e
co
m
bi
ne
d
in
th
e
w
ay
th
ey
ar
e,
or
ho
w
an
yo
ne
co
ul
d
ha
ve
th
ou
gh
to
f
th
e
pr
oo
f.
If
so
,w
e
as
k
yo
u
to
be
pa
tie
nt
.M
an
y
of
th
es
e
qu
es
tio
ns
w
ill
be
an
sw
er
ed
la
te
ri
n
th
is
bo
ok
,p
ar
tic
ul
ar
ly
in
C
ha
pt
er
3.
A
ll
of
ou
r
ex
am
pl
es
of
pr
oo
fs
in
th
is
in
tr
od
uc
tio
n
w
ill
in
vo
lv
e
pr
im
e
nu
m
-
be
rs
.
R
ec
al
l
th
at
an
in
te
ge
r
la
rg
er
th
an
1
is
sa
id
to
be
pr
im
e
if
it
ca
nn
ot
be
w
ri
tte
n
as
a
pr
od
uc
t
of
tw
o
sm
al
le
r
po
si
tiv
e
in
te
ge
rs
.F
or
ex
am
pl
e,
6
is
no
t
a
pr
im
e
nu
m
be
r,
si
nc
e
6
=
2
·3
,b
ut
7
is
a
pr
im
e
nu
m
be
r.
B
ef
or
e
w
e
ca
n
gi
ve
an
ex
am
pl
e
of
a
pr
oo
f
in
vo
lv
in
g
pr
im
e
nu
m
be
rs
,
w
e
ne
ed
to
fin
d
so
m
et
hi
ng
to
pr
ov
e
–
so
m
e
fa
ct
ab
ou
t
pr
im
e
nu
m
be
rs
w
ho
se
co
rr
ec
tn
es
s
ca
n
be
ve
ri
fie
d
w
ith
a
pr
oo
f.
So
m
et
im
es
yo
u
ca
n
fin
d
in
te
re
st
in
g
1
05
21
86
12
41
in
t
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
18
,2
00
5
17
:2
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
2
In
tr
od
uc
ti
on
pa
tte
rn
s
in
m
at
he
m
at
ic
s
ju
st
by
tr
yi
ng
ou
t
a
ca
lc
ul
at
io
n
on
a
fe
w
nu
m
be
rs
.
Fo
r
ex
am
pl
e,
co
ns
id
er
th
e
ta
bl
e
in
Fi
gu
re
1.
Fo
r
ea
ch
in
te
ge
r
n
fr
om
2
to
10
,
th
e
ta
bl
e
sh
ow
s
w
he
th
er
or
no
t
bo
th
n
an
d
2n
−
1
ar
e
pr
im
e,
an
d
a
su
rp
ri
si
ng
pa
tte
rn
em
er
ge
s.
It
ap
pe
ar
s
th
at
2n
−
1
is
pr
im
e
in
pr
ec
is
el
y
th
os
e
ca
se
s
in
w
hi
ch
n
is
pr
im
e!
Fi
gu
re
1
W
ill
th
is
pa
tte
rn
co
nt
in
ue
?
It
is
te
m
pt
in
g
to
gu
es
s
th
at
it
w
ill
,
bu
t
th
is
is
on
ly
a
gu
es
s.
M
at
he
m
at
ic
ia
ns
ca
ll
su
ch
gu
es
se
s
co
nj
ec
tu
re
s.
T
hu
s,
w
e
ha
ve
th
e
fo
llo
w
in
g
tw
o
co
nj
ec
tu
re
s:
C
on
je
ct
ur
e
1.
Su
pp
os
e
n
is
an
in
te
ge
r
la
rg
er
th
an
1
an
d
n
is
pr
im
e.
T
he
n
2n
−
1
is
pr
im
e.
C
on
je
ct
ur
e
2.
Su
pp
os
e
n
is
an
in
te
ge
r
la
rg
er
th
an
1
an
d
n
is
no
tp
ri
m
e.
T
he
n
2n
−
1
is
no
tp
ri
m
e.
U
nf
or
tu
na
te
ly
,i
fw
e
co
nt
in
ue
th
e
ta
bl
e
in
Fi
gu
re
1,
w
e
im
m
ed
ia
te
ly
fin
d
th
at
C
on
je
ct
ur
e
1
is
in
co
rr
ec
t.
It
is
ea
sy
to
ch
ec
k
th
at
11
is
pr
im
e,
bu
t
21
1
−
1
=
20
47
=
23
·8
9,
so
21
1
−
1
is
no
t
pr
im
e.
T
hu
s,
11
is
a
co
un
te
re
xa
m
pl
e
to
C
on
je
ct
ur
e
1.
T
he
ex
is
te
nc
e
of
ev
en
on
e
co
un
te
re
xa
m
pl
e
es
ta
bl
is
he
s
th
at
th
e
co
nj
ec
tu
re
is
in
co
rr
ec
t,
bu
t
it
is
in
te
re
st
in
g
to
no
te
th
at
in
th
is
ca
se
th
er
e
ar
e
m
an
y
co
un
te
re
xa
m
pl
es
.
If
w
e
co
nt
in
ue
ch
ec
ki
ng
nu
m
be
rs
up
to
30
,
w
e
fin
d
tw
o
m
or
e
co
un
te
re
xa
m
pl
es
to
C
on
je
ct
ur
e
1:
B
ot
h
23
an
d
29
ar
e
pr
im
e,
bu
t
22
3
−
1
=
8,
38
8,
60
7
=
47
·1
78
,4
81
an
d
22
9
−
1
=
53
6,
87
0,
91
1
=
2,
08
9
·
25
6,
99
9.
H
ow
ev
er
,n
o
nu
m
be
r
up
to
30
is
a
co
un
te
re
xa
m
pl
e
to
C
on
je
ct
ur
e
2.
D
o
yo
u
th
in
k
th
at
C
on
je
ct
ur
e
2
is
co
rr
ec
t?
H
av
in
g
fo
un
d
co
un
te
re
xa
m
pl
es
to
C
on
je
ct
ur
e
1,
w
e
kn
ow
th
at
th
is
co
nj
ec
tu
re
is
in
co
rr
ec
t,
bu
to
ur
fa
ilu
re
to
fin
d
a
05
21
86
12
41
in
t
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
18
,2
00
5
17
:2
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
In
tr
od
uc
ti
on
3
co
un
te
re
xa
m
pl
e
to
C
on
je
ct
ur
e
2
do
es
no
ts
ho
w
th
at
it
is
co
rr
ec
t.
Pe
rh
ap
s
th
er
e
ar
e
co
un
te
re
xa
m
pl
es
,b
ut
th
e
sm
al
le
st
on
e
is
la
rg
er
th
an
30
.C
on
tin
ui
ng
to
ch
ec
k
ex
am
pl
es
m
ig
ht
un
co
ve
r
a
co
un
te
re
xa
m
pl
e,
or
,i
f
it
do
es
n’
t,
it
m
ig
ht
in
cr
ea
se
ou
r
co
nfi
de
nc
e
in
th
e
co
nj
ec
tu
re
.B
ut
w
e
ca
n
ne
ve
r
be
su
re
th
at
th
e
co
nj
ec
tu
re
is
co
rr
ec
ti
fw
e
on
ly
ch
ec
k
ex
am
pl
es
.N
o
m
at
te
rh
ow
m
an
y
ex
am
pl
es
w
e
ch
ec
k,
th
er
e
is
al
w
ay
s
th
e
po
ss
ib
ili
ty
th
at
th
e
ne
xt
on
e
w
ill
be
th
e
fir
st
co
un
te
re
xa
m
pl
e.
T
he
only
w
ay
w
e
ca
n
be
su
re
th
at
C
on
je
ct
ur
e
2
is
co
rr
ec
ti
s
to
pr
ov
e
it.
In
fa
ct
,C
on
je
ct
ur
e
2
is
co
rr
ec
t.
H
er
e
is
a
pr
oo
f
of
th
e
co
nj
ec
tu
re
:
P
ro
of
of
C
on
je
ct
ur
e
2.
Si
nc
e
n
is
no
t
pr
im
e,
th
er
e
ar
e
po
si
tiv
e
in
te
ge
rs
a
an
d
b
su
ch
th
at
a
<
n,
b
<
n,
an
d
n
=
ab
.
L
et
x
=
2b
−
1
an
d
y
=
1
+
2b
+
22
b
+
··
·+
2(
a−
1)
b
.T
he
n
x
y
=
(2
b
−
1)
·(
1
+
2b
+
22
b
+
··
·+
2(
a−
1)
b
)
=
2b
·(
1
+
2b
+
22
b
+
··
·+
2(
a−
1)
b
)−
(1
+
2b
+
22
b
+
··
·+
2(
a−
1)
b
)
=
(2
b
+
22
b
+
23
b
+
··
·+
2a
b
)−
(1
+
2b
+
22
b
+
··
·+
2(
a−
1)
b
)
=
2a
b
−
1
=
2n
−
1.
Si
nc
e
b
<
n,
w
e
ca
n
co
nc
lu
de
th
at
x
=
2b
−
1
<
2n
−
1.
A
ls
o,
si
nc
e
ab
=
n
>
a,
it
fo
llo
w
s
th
at
b
>
1.
T
he
re
fo
re
,
x
=
2b
−
1
>
21
−
1
=
1,
so
y
<
x
y
=
2n
−
1.
T
hu
s,
w
e
ha
ve
sh
ow
n
th
at
2n
−
1
ca
n
be
w
ri
tte
n
as
th
e
pr
od
-
uc
to
f
tw
o
po
si
tiv
e
in
te
ge
rs
x
an
d
y,
bo
th
of
w
hi
ch
ar
e
sm
al
le
r
th
an
2n
−
1,
so
2n
−
1
is
no
tp
ri
m
e.
�
N
ow
th
at
th
e
co
nj
ec
tu
re
ha
s
be
en
pr
ov
en
,
w
e
ca
n
ca
ll
it
a
th
eo
re
m
.
D
on
’t
w
or
ry
if
yo
u
fin
d
th
e
pr
oo
f
so
m
ew
ha
t
m
ys
te
ri
ou
s.
W
e’
ll
re
tu
rn
to
it
ag
ai
n
at
th
e
en
d
of
C
ha
pt
er
3
to
an
al
yz
e
ho
w
it
w
as
co
ns
tr
uc
te
d.
Fo
r
th
e
m
om
en
t,
th
e
m
os
t
im
po
rt
an
t
po
in
t
to
un
de
rs
ta
nd
is
th
at
if
n
is
an
y
in
te
ge
r
la
rg
er
th
an
1
th
at
ca
n
be
w
ri
tte
n
as
a
pr
od
uc
to
f
tw
o
sm
al
le
r
po
si
tiv
e
in
te
ge
rs
a
an
d
b,
th
en
th
e
pr
oo
f
gi
ve
s
a
m
et
ho
d
(a
dm
itt
ed
ly
,a
so
m
ew
ha
tm
ys
te
ri
ou
s
on
e)
of
w
ri
tin
g
2n
−
1
as
a
pr
od
uc
t
of
tw
o
sm
al
le
r
po
si
tiv
e
in
te
ge
rs
x
an
d
y.
T
hu
s,
if
n
is
no
t
pr
im
e,
th
en
2n
−
1
m
us
t
al
so
no
t
be
pr
im
e.
Fo
r
ex
am
pl
e,
su
pp
os
e
n
=
12
,s
o
2n
−
1
=
40
95
.
Si
nc
e
12
=
3
·4
,w
e
co
ul
d
ta
ke
a
=
3
an
d
b
=
4
in
th
e
pr
oo
f.
T
he
n
ac
co
rd
in
g
to
th
e
fo
rm
ul
as
fo
r
x
an
d
y
gi
ve
n
in
th
e
pr
oo
f,
w
e
w
ou
ld
ha
ve
x
=
2b
−
1
=
24
−
1
=
15
,
an
d
y
=
1
+
2b
+
22
b
+
··
·+
2(
a−
1)
b
=
1
+
24
+
28
=
27
3.
A
nd
,
ju
st
as
th
e
fo
rm
ul
as
in
th
e
pr
oo
f
pr
ed
ic
t,
w
e
ha
ve
x
y
=
15
·2
73
=
40
95
=
2n
−
1.
O
f
co
ur
se
,t
he
re
ar
e
ot
he
r
w
ay
s
of
fa
ct
or
in
g
12
in
to
a
pr
od
uc
to
ft
w
o
sm
al
le
ri
nt
eg
er
s,
an
d
th
es
e
m
ig
ht
le
ad
to
ot
he
rw
ay
s
of
05
21
86
12
41
in
t
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
18
,2
00
5
17
:2
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
4
In
tr
od
uc
ti
on
fa
ct
or
in
g
40
95
.F
or
ex
am
pl
e,
si
nc
e
12
=
2
·6
,w
e
co
ul
d
us
e
th
e
va
lu
es
a
=
2
an
d
b
=
6.
T
ry
co
m
pu
tin
g
th
e
co
rr
es
po
nd
in
g
va
lu
es
of
x
an
d
y
an
d
m
ak
e
su
re
th
ei
r
pr
od
uc
ti
s
40
95
.
A
lth
ou
gh
w
e
al
re
ad
y
kn
ow
th
at
C
on
je
ct
ur
e
1
is
in
co
rr
ec
t,
th
er
e
ar
e
st
ill
in
te
r-
es
tin
g
qu
es
tio
ns
w
e
ca
n
as
k
ab
ou
ti
t.
If
w
e
co
nt
in
ue
ch
ec
ki
ng
pr
im
e
nu
m
be
rs
n
to
se
e
if
2n
−
1
is
pr
im
e,
w
ill
w
e
co
nt
in
ue
to
fin
d
co
un
te
re
xa
m
pl
es
to
th
e
co
nj
ec
tu
re
–
ex
am
pl
es
fo
rw
hi
ch
2n
−
1
is
no
tp
ri
m
e?
W
ill
w
e
co
nt
in
ue
to
fin
d
ex
am
pl
es
fo
r
w
hi
ch
2n
−
1
is
pr
im
e?
If
th
er
e
w
er
e
on
ly
fin
ite
ly
m
an
y
pr
im
e
nu
m
be
rs
,t
he
n
w
e
m
ig
ht
be
ab
le
to
in
ve
st
ig
at
e
th
es
e
qu
es
tio
ns
by
si
m
pl
y
ch
ec
k-
in
g
2n
−
1
fo
re
ve
ry
pr
im
e
nu
m
be
rn
.B
ut
in
fa
ct
th
er
e
ar
e
in
fin
ite
ly
m
an
y
pr
im
e
nu
m
be
rs
.
E
uc
lid
(c
ir
ca
35
0
b.
c.
)
ga
ve
a
pr
oo
f
of
th
is
fa
ct
in
B
oo
k
IX
of
hi
s
E
le
m
en
ts
.H
is
pr
oo
f
is
on
e
of
th
e
m
os
tf
am
ou
s
in
al
lo
f
m
at
he
m
at
ic
s:
T
he
or
em
3.
T
he
re
ar
e
in
fin
it
el
y
m
an
y
pr
im
e
nu
m
be
rs
.
P
ro
of
.
Su
pp
os
e
th
er
e
ar
e
on
ly
fin
ite
ly
m
an
y
pr
im
e
nu
m
be
rs
.L
et
p 1
,
p 2
,
..
.,
p n
be
a
lis
t
of
al
l
pr
im
e
nu
m
be
rs
.
L
et
m
=
p 1
p 2
··
·p
n
+
1.
N
ot
e
th
at
m
is
no
t
di
vi
si
bl
e
by
p 1
,s
in
ce
di
vi
di
ng
m
by
p 1
gi
ve
s
a
qu
ot
ie
nt
of
p 2
p 3
··
·p
n
an
d
a
re
m
ai
nd
er
of
1.
Si
m
ila
rl
y,
m
is
no
td
iv
is
ib
le
by
an
y
of
p 2
,
p 3
,
..
.,
p n
.
W
e
no
w
us
e
th
e
fa
ct
th
at
ev
er
y
in
te
ge
r
la
rg
er
th
an
1
is
ei
th
er
pr
im
e
or
ca
n
be
w
ri
tte
n
as
a
pr
od
uc
to
f
pr
im
es
.(
W
e’
ll
se
e
a
pr
oo
f
of
th
is
fa
ct
in
C
ha
pt
er
6.
)
C
le
ar
ly
m
is
la
rg
er
th
an
1,
so
m
is
ei
th
er
pr
im
e
or
a
pr
od
uc
to
fp
ri
m
es
.S
up
po
se
fir
st
th
at
m
is
pr
im
e.
N
ot
e
th
at
m
is
la
rg
er
th
an
al
l
of
th
e
nu
m
be
rs
in
th
e
lis
t
p 1
,
p 2
,
..
.,
p n
,
so
w
e’
ve
fo
un
d
a
pr
im
e
nu
m
be
r
no
t
in
th
is
lis
t.
B
ut
th
is
co
nt
ra
di
ct
s
ou
r
as
su
m
pt
io
n
th
at
th
is
w
as
a
lis
to
f
al
lp
ri
m
e
nu
m
be
rs
.
N
ow
su
pp
os
e
m
is
a
pr
od
uc
t
of
pr
im
es
.
L
et
q
be
on
e
of
th
e
pr
im
es
in
th
is
pr
od
uc
t.
T
he
n
m
is
di
vi
si
bl
e
by
q.
B
ut
w
e’
ve
al
re
ad
y
se
en
th
at
m
is
no
td
iv
is
ib
le
by
an
y
of
th
e
nu
m
be
rs
in
th
e
lis
t
p 1
,
p 2
,
..
.,
p n
,
so
on
ce
ag
ai
n
w
e
ha
ve
a
co
nt
ra
di
ct
io
n
w
ith
th
e
as
su
m
pt
io
n
th
at
th
is
lis
ti
nc
lu
de
d
al
lp
ri
m
e
nu
m
be
rs
.
Si
nc
e
th
e
as
su
m
pt
io
n
th
at
th
er
e
ar
e
fin
ite
ly
m
an
y
pr
im
e
nu
m
be
rs
ha
s
le
d
to
a
co
nt
ra
di
ct
io
n,
th
er
e
m
us
tb
e
in
fin
ite
ly
m
an
y
pr
im
e
nu
m
be
rs
.
�
O
nc
e
ag
ai
n,
yo
u
sh
ou
ld
no
tb
e
co
nc
er
ne
d
if
so
m
e
as
pe
ct
s
of
th
is
pr
oo
fs
ee
m
m
ys
te
ri
ou
s.
A
ft
er
yo
u’
ve
re
ad
C
ha
pt
er
3
yo
u’
ll
be
be
tte
rp
re
pa
re
d
to
un
de
rs
ta
nd
th
e
pr
oo
f
in
de
ta
il.
W
e’
ll
re
tu
rn
to
th
is
pr
oo
f
th
en
an
d
an
al
yz
e
its
st
ru
ct
ur
e.
W
e
ha
ve
se
en
th
at
if
n
is
no
tp
ri
m
e
th
en
2n
−
1
ca
nn
ot
be
pr
im
e,
bu
ti
f
n
is
pr
im
e
th
en
2n
−
1
ca
n
be
ei
th
er
pr
im
e
or
no
tp
ri
m
e.
B
ec
au
se
th
er
e
ar
e
in
fin
ite
ly
m
an
y
pr
im
e
nu
m
be
rs
,
th
er
e
ar
e
in
fin
ite
ly
m
an
y
nu
m
be
rs
of
th
e
fo
rm
2n
−
1
th
at
,b
as
ed
on
w
ha
t
w
e
kn
ow
so
fa
r,
m
ig
ht
be
pr
im
e.
B
ut
ho
w
m
an
y
of
th
em
ar
e
pr
im
e?
05
21
86
12
41
in
t
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
18
,2
00
5
17
:2
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
In
tr
od
uc
ti
on
5
Pr
im
e
nu
m
be
rs
of
th
e
fo
rm
2n
−
1
ar
e
ca
lle
d
M
er
se
nn
e
pr
im
es
,a
ft
er
Fa
th
er
M
ar
in
M
er
se
nn
e
(1
58
8–
16
47
),
a
Fr
en
ch
m
on
k
an
d
sc
ho
la
r
w
ho
st
ud
ie
d
th
es
e
nu
m
be
rs
.
A
lth
ou
gh
m
an
y
M
er
se
nn
e
pr
im
es
ha
ve
be
en
fo
un
d,
it
is
st
ill
no
t
kn
ow
n
if
th
er
e
ar
e
in
fin
ite
ly
m
an
y
of
th
em
.M
an
y
of
th
e
la
rg
es
tk
no
w
n
pr
im
e
nu
m
be
rs
ar
e
M
er
se
nn
e
pr
im
es
.
A
s
of
th
is
w
ri
tin
g
(A
pr
il
20
05
),
th
e
la
rg
es
t
kn
ow
n
pr
im
e
nu
m
be
r
is
th
e
M
er
se
nn
e
pr
im
e
22
5,
96
4,
95
1
−
1,
a
nu
m
be
r
w
ith
7,
81
6,
23
0
di
gi
ts
.
M
er
se
nn
e
pr
im
es
ar
e
re
la
te
d
to
pe
rf
ec
t
nu
m
be
rs
,t
he
su
bj
ec
t
of
an
ot
he
r
fa
-
m
ou
s
un
so
lv
ed
pr
ob
le
m
of
m
at
he
m
at
ic
s.
A
po
si
tiv
e
in
te
ge
r
n
is
sa
id
to
be
pe
rf
ec
ti
fn
is
eq
ua
lt
o
th
e
su
m
of
al
lp
os
itive
in
te
ge
rs
sm
al
le
rt
ha
n
n
th
at
di
vi
de
n.
(F
or
an
y
tw
o
in
te
ge
rs
m
an
d
n,
w
e
sa
y
th
at
m
di
vi
de
s
n
if
n
is
di
vi
si
bl
e
by
m
;
in
ot
he
rw
or
ds
,i
ft
he
re
is
an
in
te
ge
rq
su
ch
th
at
n
=
q
m
.)
Fo
re
xa
m
pl
e,
th
e
on
ly
po
si
tiv
e
in
te
ge
rs
sm
al
le
rt
ha
n
6
th
at
di
vi
de
6
ar
e
1,
2,
an
d
3,
an
d
1
+
2
+
3
=
6.
T
hu
s,
6
is
a
pe
rf
ec
tn
um
be
r.
T
he
ne
xt
sm
al
le
st
pe
rf
ec
tn
um
be
ri
s2
8.
(Y
ou
sh
ou
ld
ch
ec
k
fo
r
yo
ur
se
lf
th
at
28
is
pe
rf
ec
tb
y
fin
di
ng
al
lt
he
po
si
tiv
e
in
te
ge
rs
sm
al
le
r
th
an
28
th
at
di
vi
de
28
an
d
ad
di
ng
th
em
up
.)
E
uc
lid
pr
ov
ed
th
at
if
2n
−
1
is
pr
im
e,
th
en
2n
−1
(2
n
−
1)
is
pe
rf
ec
t.
T
hu
s,
ev
er
y
M
er
se
nn
e
pr
im
e
gi
ve
s
ri
se
to
a
pe
rf
ec
t
nu
m
be
r.
Fu
rt
he
rm
or
e,
ab
ou
t
20
00
ye
ar
s
af
te
r
E
uc
lid
’s
pr
oo
f,
th
e
Sw
is
s
m
at
he
m
at
ic
ia
n
L
eo
nh
ar
d
E
ul
er
(1
70
7–
17
83
),
th
e
m
os
t
pr
ol
ifi
c
m
at
he
m
at
ic
ia
n
in
hi
st
or
y,
pr
ov
ed
th
at
ev
er
y
ev
en
pe
rf
ec
tn
um
be
ra
ri
se
s
in
th
is
w
ay
.(
Fo
re
xa
m
pl
e,
no
te
th
at
6
=
21
(2
2
−
1)
an
d
28
=
22
(2
3
−
1)
.)
B
ec
au
se
it
is
no
t
kn
ow
n
if
th
er
e
ar
e
in
fin
ite
ly
m
an
y
M
er
se
nn
e
pr
im
es
,i
ti
s
al
so
no
tk
no
w
n
if
th
er
e
ar
e
in
fin
ite
ly
m
an
y
ev
en
pe
rf
ec
t
nu
m
be
rs
.I
ti
s
al
so
no
tk
no
w
n
if
th
er
e
ar
e
an
y
od
d
pe
rf
ec
tn
um
be
rs
.
A
lth
ou
gh
th
er
e
ar
e
in
fin
ite
ly
m
an
y
pr
im
e
nu
m
be
rs
,
th
e
pr
im
es
th
in
ou
t
as
w
e
lo
ok
at
la
rg
er
an
d
la
rg
er
nu
m
be
rs
.
Fo
r
ex
am
pl
e,
th
er
e
ar
e
25
pr
im
es
be
-
tw
ee
n
1
an
d
10
0,
16
pr
im
es
be
tw
ee
n
10
00
an
d
11
00
,
an
d
on
ly
si
x
pr
im
es
be
tw
ee
n
1,
00
0,
00
0
an
d
1,
00
0,
10
0.
A
s
ou
rl
as
ti
nt
ro
du
ct
or
y
ex
am
pl
e
of
a
pr
oo
f,
w
e
sh
ow
th
at
th
er
e
ar
e
lo
ng
st
re
tc
he
s
of
co
ns
ec
ut
iv
e
po
si
tiv
e
in
te
ge
rs
co
n-
ta
in
in
g
no
pr
im
es
at
al
l.
In
th
is
pr
oo
f,
w
e’
ll
us
e
th
e
fo
llo
w
in
g
te
rm
in
ol
og
y:
Fo
r
an
y
po
si
tiv
e
in
te
ge
r
n,
th
e
pr
od
uc
t
of
al
l
in
te
ge
rs
fr
om
1
to
n
is
ca
lle
d
n
fa
ct
or
ia
l
an
d
is
de
no
te
d
n!
.T
hu
s,
n!
=
1
·2
·3
··
·n
.A
s
w
ith
ou
r
pr
ev
io
us
tw
o
pr
oo
fs
,
w
e’
ll
re
tu
rn
to
th
is
pr
oo
f
at
th
e
en
d
of
C
ha
pt
er
3
to
an
al
yz
e
its
st
ru
ct
ur
e.
T
he
or
em
4.
Fo
r
ev
er
y
po
si
ti
ve
in
te
ge
r
n,
th
er
e
is
a
se
qu
en
ce
of
n
co
ns
ec
ut
iv
e
po
si
ti
ve
in
te
ge
rs
co
nt
ai
ni
ng
no
pr
im
es
.
P
ro
of
.
Su
pp
os
e
n
is
a
po
si
tiv
e
in
te
ge
r.
L
et
x
=
(n
+
1)
!+
2.
W
e
w
ill
sh
ow
th
at
no
ne
of
th
e
nu
m
be
rs
x,
x
+
1,
x
+
2,
..
.,
x
+
(n
−
1)
is
pr
im
e.
Si
nc
e
th
is
is
a
se
qu
en
ce
of
n
co
ns
ec
ut
iv
e
po
si
tiv
e
in
te
ge
rs
,t
hi
s
w
ill
pr
ov
e
th
e
th
eo
re
m
.
05
21
86
12
41
in
t
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
18
,2
00
5
17
:2
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
6
In
tr
od
uc
ti
on
To
se
e
th
at
x
is
no
tp
ri
m
e,
no
te
th
at
x
=
1
·2
·3
·4
··
·(n
+
1)
+
2
=
2
·(
1
·3
·4
··
·(n
+
1)
+
1)
.
T
hu
s,
x
ca
n
be
w
ri
tte
n
as
a
pr
od
uc
to
f
tw
o
sm
al
le
r
po
si
tiv
e
in
te
ge
rs
,s
o
x
is
no
tp
ri
m
e.
Si
m
ila
rl
y,
w
e
ha
ve x
+
1
=
1
·2
·3
·4
··
·(n
+
1)
+
3
=
3
·(
1
·2
·4
··
·(n
+
1)
+
1)
,
so
x
+
1
is
al
so
no
t
pr
im
e.
In
ge
ne
ra
l,
co
ns
id
er
an
y
nu
m
be
r
x
+
i,
w
he
re
0
≤
i
≤
n
−
1.
T
he
n
w
e
ha
ve
x
+
i
=
1
·2
·3
·4
··
·(n
+
1)
+
(i
+
2)
=
(i
+
2)
·(
1
·2
·3
··
·(i
+
1)
·(
i
+
3)
··
·(n
+
1)
+
1)
,
so
x
+
i
is
no
tp
ri
m
e.
�
T
he
or
em
4
sh
ow
s
th
at
th
er
e
ar
e
so
m
et
im
es
lo
ng
st
re
tc
he
s
be
tw
ee
n
on
e
pr
im
e
an
d
th
e
ne
xt
pr
im
e.
B
ut
pr
im
es
al
so
so
m
et
im
es
oc
cu
r
cl
os
e
to
ge
th
er
.S
in
ce
2
is
th
e
on
ly
ev
en
pr
im
e
nu
m
be
r,
th
e
on
ly
pa
ir
of
co
ns
ec
ut
iv
e
in
te
ge
rs
th
at
ar
e
bo
th
pr
im
e
is
2
an
d
3.
B
ut
th
er
e
ar
e
lo
ts
of
pa
ir
s
of
pr
im
es
th
at
di
ff
er
by
on
ly
tw
o,
fo
re
xa
m
pl
e,
5
an
d
7,
29
an
d
31
,a
nd
79
49
an
d
79
51
.S
uc
h
pa
ir
s
of
pr
im
es
ar
e
ca
lle
d
tw
in
pr
im
es
.I
ti
s
no
tk
no
w
n
w
he
th
er
th
er
e
ar
e
in
fin
ite
ly
m
an
y
tw
in
pr
im
es
.
E
xe
rc
is
es
∗ 1
.
(a
)
Fa
ct
or
21
5
−
1
=
32
,7
67
in
to
a
pr
od
uc
to
ft
w
o
sm
al
le
rp
os
iti
ve
in
te
ge
rs
.
(b
)
Fi
nd
an
in
te
ge
r
x
su
ch
th
at
1
<
x
<
23
27
67
−
1
an
d
23
27
67
−
1
is
di
vi
s-
ib
le
by
x.
2.
M
ak
e
so
m
e
co
nj
ec
tu
re
s
ab
ou
tt
he
va
lu
es
of
n
fo
r
w
hi
ch
3n
−
1
is
pr
im
e
or
th
e
va
lu
es
of
n
fo
r
w
hi
ch
3n
−
2n
is
pr
im
e.
(Y
ou
m
ig
ht
st
ar
t
by
m
ak
in
g
a
ta
bl
e
si
m
ila
r
to
Fi
gu
re
1.
)
∗ 3
.
T
he
pr
oo
fo
fT
he
or
em
3
gi
ve
sa
m
et
ho
d
fo
rfi
nd
in
g
a
pr
im
e
nu
m
be
rd
if
fe
re
nt
fr
om
an
y
in
a
gi
ve
n
lis
to
f
pr
im
e
nu
m
be
rs
.
(a
)
U
se
th
is
m
et
ho
d
to
fin
d
a
pr
im
e
di
ff
er
en
tf
ro
m
2,
3,
5,
an
d
7.
(b
)
U
se
th
is
m
et
ho
d
to
fin
d
a
pr
im
e
di
ff
er
en
tf
ro
m
2,
5,
an
d
11
.
4.
Fi
nd
fiv
e
co
ns
ec
ut
iv
e
in
te
ge
rs
th
at
ar
e
no
tp
ri
m
e.
05
21
86
12
41
in
t
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
18
,2
00
5
17
:2
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
In
tr
od
uc
ti
on
7
5.
U
se
th
e
ta
bl
e
in
Fi
gu
re
1
an
d
th
e
di
sc
us
si
on
on
p.
5
to
fin
d
tw
o
m
or
e
pe
rf
ec
t
nu
m
be
rs
.
6.
T
he
se
qu
en
ce
3,
5,
7
is
a
lis
to
f
th
re
e
pr
im
e
nu
m
be
rs
su
ch
th
at
ea
ch
pa
ir
of
ad
ja
ce
nt
nu
m
be
rs
in
th
e
lis
td
if
fe
rb
y
tw
o.
A
re
th
er
e
an
y
m
or
e
su
ch
“t
ri
pl
et
pr
im
es
”?
05
21
86
12
41
c0
1
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
1
Se
nt
en
tia
lL
og
ic
1.
1.
D
ed
uc
ti
ve
R
ea
so
ni
ng
an
d
L
og
ic
al
C
on
ne
ct
iv
es
A
s
w
e
sa
w
in
th
e
in
tr
od
uc
tio
n,
pr
oo
fs
pl
ay
a
ce
nt
ra
lr
ol
e
in
m
at
he
m
at
ic
s,
an
d
de
du
ct
iv
e
re
as
on
in
g
is
th
e
fo
un
da
tio
n
on
w
hi
ch
pr
oo
fs
ar
e
ba
se
d.
T
he
re
fo
re
,
w
e
be
gi
n
ou
r
st
ud
y
of
m
at
he
m
at
ic
al
re
as
on
in
g
an
d
pr
oo
fs
by
ex
am
in
in
g
ho
w
de
du
ct
iv
e
re
as
on
in
g
w
or
ks
.
E
xa
m
pl
e
1.
1.
1.
H
er
e
ar
e
th
re
e
ex
am
pl
es
of
de
du
ct
iv
e
re
as
on
in
g:
1.
It
w
ill
ei
th
er
ra
in
or
sn
ow
to
m
or
ro
w
.
It
’s
to
o
w
ar
m
fo
r
sn
ow
.
T
he
re
fo
re
,i
tw
ill
ra
in
.
2.
If
to
da
y
is
Su
nd
ay
,t
he
n
I
do
n’
th
av
e
to
go
to
w
or
k
to
da
y.
To
da
y
is
Su
nd
ay
.
T
he
re
fo
re
,I
do
n’
th
av
e
to
go
to
w
or
k
to
da
y.
3.
I
w
ill
go
to
w
or
k
ei
th
er
to
m
or
ro
w
or
to
da
y.
I’
m
go
in
g
to
st
ay
ho
m
e
to
da
y.
T
he
re
fo
re
,I
w
ill
go
to
w
or
k
to
m
or
ro
w
.
In
ea
ch
ca
se
,
w
e
ha
ve
ar
ri
ve
d
at
a
co
nc
lu
si
on
fr
om
th
e
as
su
m
pt
io
n
th
at
so
m
e
ot
he
r
st
at
em
en
ts
,c
al
le
d
pr
em
is
es
,a
re
tr
ue
.F
or
ex
am
pl
e,
th
e
pr
em
is
es
in
ar
gu
m
en
t
3
ar
e
th
e
st
at
em
en
ts
“I
w
ill
go
to
w
or
k
ei
th
er
to
m
or
ro
w
or
to
da
y”
an
d
“I
’m
go
in
g
to
st
ay
ho
m
e
to
da
y.
”
T
he
co
nc
lu
si
on
is
“I
w
ill
go
to
w
or
k
to
m
or
ro
w
,”
an
d
it
se
em
s
to
be
fo
rc
ed
on
us
so
m
eh
ow
by
th
e
pr
em
is
es
.
B
ut
is
th
is
co
nc
lu
si
on
re
al
lyco
rr
ec
t?
A
ft
er
al
l,
is
n’
ti
tp
os
si
bl
e
th
at
I’
ll
st
ay
ho
m
e
to
da
y,
an
d
th
en
w
ak
e
up
si
ck
to
m
or
ro
w
an
d
en
d
up
st
ay
in
g
ho
m
e
ag
ai
n?
If
th
at
ha
pp
en
ed
,t
he
co
nc
lu
si
on
w
ou
ld
tu
rn
ou
tt
o
be
fa
ls
e.
B
ut
no
tic
e
th
at
in
th
at
ca
se
th
e
fir
st
pr
em
is
e,
w
hi
ch
sa
id
th
at
I
w
ou
ld
go
to
w
or
k
ei
th
er
to
m
or
ro
w
8
05
21
86
12
41
c0
1
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
D
ed
uc
ti
ve
R
ea
so
ni
ng
an
d
L
og
ic
al
C
on
ne
ct
iv
es
9
or
to
da
y,
w
ou
ld
be
fa
ls
e
as
w
el
l!
A
lth
ou
gh
w
e
ha
ve
no
gu
ar
an
te
e
th
at
th
e
co
nc
lu
si
on
is
tr
ue
,
it
ca
n
on
ly
be
fa
ls
e
if
at
le
as
t
on
e
of
th
e
pr
em
is
es
is
al
so
fa
ls
e.
If
bo
th
pr
em
is
es
ar
e
tr
ue
,w
e
ca
n
be
su
re
th
at
th
e
co
nc
lu
si
on
is
al
so
tr
ue
.
T
hi
s
is
th
e
se
ns
e
in
w
hi
ch
th
e
co
nc
lu
si
on
is
fo
rc
ed
on
us
by
th
e
pr
em
is
es
,a
nd
th
is
is
th
e
st
an
da
rd
w
e
w
ill
us
e
to
ju
dg
e
th
e
co
rr
ec
tn
es
s
of
de
du
ct
iv
e
re
as
on
in
g.
W
e
w
ill
sa
y
th
at
an
ar
gu
m
en
ti
s
va
li
d
if
th
e
pr
em
is
es
ca
nn
ot
al
lb
e
tr
ue
w
ith
ou
t
th
e
co
nc
lu
si
on
be
in
g
tr
ue
as
w
el
l.
A
ll
th
re
e
of
th
e
ar
gu
m
en
ts
in
ou
r
ex
am
pl
e
ar
e
va
lid
ar
gu
m
en
ts
.
H
er
e’
s
an
ex
am
pl
e
of
an
in
va
lid
de
du
ct
iv
e
ar
gu
m
en
t:
E
ith
er
th
e
bu
tle
r
is
gu
ilt
y
or
th
e
m
ai
d
is
gu
ilt
y.
E
ith
er
th
e
m
ai
d
is
gu
ilt
y
or
th
e
co
ok
is
gu
ilt
y.
T
he
re
fo
re
,e
ith
er
th
e
bu
tle
r
is
gu
ilt
y
or
th
e
co
ok
is
gu
ilt
y.
T
he
ar
gu
m
en
t
is
in
va
lid
be
ca
us
e
th
e
co
nc
lu
si
on
co
ul
d
be
fa
ls
e
ev
en
if
bo
th
pr
em
is
es
ar
e
tr
ue
.F
or
ex
am
pl
e,
if
th
e
m
ai
d
w
er
e
gu
ilt
y,
bu
tt
he
bu
tle
r
an
d
th
e
co
ok
w
er
e
bo
th
in
no
ce
nt
,t
he
n
bo
th
pr
em
is
es
w
ou
ld
be
tr
ue
an
d
th
e
co
nc
lu
si
on
w
ou
ld
be
fa
ls
e.
W
e
ca
n
le
ar
n
so
m
et
hi
ng
ab
ou
t
w
ha
t
m
ak
es
an
ar
gu
m
en
t
va
lid
by
co
m
pa
r-
in
g
th
e
th
re
e
ar
gu
m
en
ts
in
E
xa
m
pl
e
1.
1.
1.
O
n
th
e
su
rf
ac
e
it
m
ig
ht
se
em
th
at
ar
gu
m
en
ts
2
an
d
3
ha
ve
th
e
m
os
t
in
co
m
m
on
,
be
ca
us
e
th
ey
’r
e
bo
th
ab
ou
t
th
e
sa
m
e
su
bj
ec
t:
at
te
nd
an
ce
at
w
or
k.
B
ut
in
te
rm
s
of
th
e
re
as
on
in
g
us
ed
,
ar
gu
m
en
ts
1
an
d
3
ar
e
th
e
m
os
t
si
m
ila
r.
T
he
y
bo
th
in
tr
od
uc
e
tw
o
po
ss
ib
ili
-
tie
s
in
th
e
fir
st
pr
em
is
e,
ru
le
ou
tt
he
se
co
nd
on
e
w
ith
th
e
se
co
nd
pr
em
is
e,
an
d
th
en
co
nc
lu
de
th
at
th
e
fir
st
po
ss
ib
ili
ty
m
us
t
be
th
e
ca
se
.I
n
ot
he
r
w
or
ds
,b
ot
h
ar
gu
m
en
ts
ha
ve
th
e
fo
rm
:
P
or
Q
.
no
tQ
.
T
he
re
fo
re
,P
.
It
is
th
is
fo
rm
,
an
d
no
t
th
e
su
bj
ec
t
m
at
te
r,
th
at
m
ak
es
th
es
e
ar
gu
m
en
ts
va
lid
.
Y
ou
ca
n
se
e
th
at
ar
gu
m
en
t1
ha
s
th
is
fo
rm
by
th
in
ki
ng
of
th
e
le
tte
rP
as
st
an
di
ng
fo
r
th
e
st
at
em
en
t“
It
w
ill
ra
in
to
m
or
ro
w
,”
an
d
Q
as
st
an
di
ng
fo
r
“I
tw
ill
sn
ow
to
m
or
ro
w
.”
Fo
r
ar
gu
m
en
t3
,P
w
ou
ld
be
“I
w
ill
go
to
w
or
k
to
m
or
ro
w
,”
an
d
Q
w
ou
ld
be
“I
w
ill
go
to
w
or
k
to
da
y.
”
R
ep
la
ci
ng
ce
rt
ai
n
st
at
em
en
ts
in
ea
ch
ar
gu
m
en
t
w
ith
le
tte
rs
,
as
w
e
ha
ve
in
st
at
in
g
th
e
fo
rm
of
ar
gu
m
en
ts
1
an
d
3,
ha
s
tw
o
ad
va
nt
ag
es
.F
ir
st
,i
t
ke
ep
s
us
fr
om
be
in
g
di
st
ra
ct
ed
by
as
pe
ct
so
ft
he
ar
gu
m
en
ts
th
at
do
n’
ta
ff
ec
tt
he
ir
va
lid
ity
.
Y
ou
do
n’
tn
ee
d
to
kn
ow
an
yt
hi
ng
ab
ou
tw
ea
th
er
fo
re
ca
st
in
g
or
w
or
k
ha
bi
ts
to
re
co
gn
iz
e
th
at
ar
gu
m
en
ts
1
an
d
3
ar
e
va
lid
.T
ha
t’s
be
ca
us
e
bo
th
ar
gu
m
en
ts
ha
ve
th
e
fo
rm
sh
ow
n
ea
rl
ie
r,
an
d
yo
u
ca
n
te
ll
th
at
th
is
ar
gu
m
en
tf
or
m
is
va
lid
w
ith
ou
t
05
21
86
12
41
c0
1
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
10
Se
nt
en
ti
al
L
og
ic
ev
en
kn
ow
in
g
w
ha
t
P
an
d
Q
st
an
d
fo
r.
If
yo
u
do
n’
t
be
lie
ve
th
is
,c
on
si
de
r
th
e
fo
llo
w
in
g
ar
gu
m
en
t:
E
ith
er
th
e
fr
am
ge
rw
id
ge
ti
s
m
is
fir
in
g,
or
th
e
w
ro
m
pa
lm
ec
ha
ni
sm
is
ou
to
f
al
ig
nm
en
t.
I’
ve
ch
ec
ke
d
th
e
al
ig
nm
en
t
of
th
e
w
ro
m
pa
l
m
ec
ha
ni
sm
,
an
d
it’
s
fin
e.
T
he
re
fo
re
,t
he
fr
am
ge
r
w
id
ge
ti
s
m
is
fir
in
g.
If
a
m
ec
ha
ni
c
ga
ve
th
is
ex
pl
an
at
io
n
af
te
r
ex
am
in
in
g
yo
ur
ca
r,
yo
u
m
ig
ht
st
ill
be
m
ys
tifi
ed
ab
ou
tw
hy
th
e
ca
rw
on
’t
st
ar
t,
bu
ty
ou
’d
ha
ve
no
tr
ou
bl
e
fo
llo
w
in
g
hi
s
lo
gi
c!
Pe
rh
ap
s
m
or
e
im
po
rt
an
t,
ou
r
an
al
ys
is
of
th
e
fo
rm
s
of
ar
gu
m
en
ts
1
an
d
3
m
ak
es
cl
ea
r
w
ha
ti
s
im
po
rt
an
ti
n
de
te
rm
in
in
g
th
ei
r
va
lid
ity
:t
he
w
or
ds
or
an
d
no
t.
In
m
os
td
ed
uc
tiv
e
re
as
on
in
g,
an
d
in
pa
rt
ic
ul
ar
in
m
at
he
m
at
ic
al
re
as
on
in
g,
th
e
m
ea
ni
ng
s
of
ju
st
a
fe
w
w
or
ds
gi
ve
us
th
e
ke
y
to
un
de
rs
ta
nd
in
g
w
ha
tm
ak
es
a
pi
ec
e
of
re
as
on
in
g
va
lid
or
in
va
lid
.
(W
hi
ch
ar
e
th
e
im
po
rt
an
t
w
or
ds
in
ar
-
gu
m
en
t
2
in
E
xa
m
pl
e
1.
1.
1?
)
T
he
fir
st
fe
w
ch
ap
te
rs
of
th
is
bo
ok
ar
e
de
vo
te
d
to
st
ud
yi
ng
th
os
e
w
or
ds
an
d
ho
w
th
ey
ar
e
us
ed
in
m
at
he
m
at
ic
al
w
ri
tin
g
an
d
re
as
on
in
g.
In
th
is
ch
ap
te
r,
w
e’
ll
co
nc
en
tr
at
e
on
w
or
ds
us
ed
to
co
m
bi
ne
st
at
em
en
ts
to
fo
rm
m
or
e
co
m
pl
ex
st
at
em
en
ts
.W
e’
ll
co
nt
in
ue
to
us
e
le
tte
rs
to
st
an
d
fo
rs
ta
te
-
m
en
ts
,b
ut
on
ly
fo
ru
na
m
bi
gu
ou
s
st
at
em
en
ts
th
at
ar
e
ei
th
er
tr
ue
or
fa
ls
e.
Q
ue
s-
tio
ns
,e
xc
la
m
at
io
ns
,a
nd
va
gu
e
st
at
em
en
ts
w
ill
no
t
be
al
lo
w
ed
.I
t
w
ill
al
so
be
us
ef
ul
to
us
e
sy
m
bo
ls
,s
om
et
im
es
ca
lle
d
co
nn
ec
ti
ve
sy
m
bo
ls
,t
o
st
an
d
fo
rs
om
e
of
th
e
w
or
ds
us
ed
to
co
m
bi
ne
st
at
em
en
ts
.H
er
e
ar
e
ou
r
fir
st
th
re
e
co
nn
ec
tiv
e
sy
m
bo
ls
an
d
th
e
w
or
ds
th
ey
st
an
d
fo
r:
Sy
m
bo
l
M
ea
ni
ng
∨
or
∧
an
d
¬
no
t
T
hu
s,
if
P
an
d
Q
st
an
d
fo
r
tw
o
st
at
em
en
ts
,t
he
n
w
e’
ll
w
ri
te
P
∨
Q
to
st
an
d
fo
r
th
e
st
at
em
en
t
“P
or
Q
,”
P
∧
Q
fo
r
“P
an
d
Q
,”
an
d
¬P
fo
r
“n
ot
P
”
or
“P
is
fa
ls
e.
”
T
he
st
at
em
en
t
P
∨
Q
is
so
m
et
im
es
ca
lle
d
th
e
di
sj
un
ct
io
n
of
P
an
d
Q
,
P
∧
Q
is
ca
lle
d
th
e
co
nj
un
ct
io
n
of
P
an
d
Q
,
an
d
¬P
is
ca
lle
d
th
e
ne
ga
ti
on
of
P
.
E
xa
m
pl
e
1.
1.
2.
A
na
ly
ze
th
e
lo
gi
ca
lf
or
m
s
of
th
e
fo
llo
w
in
g
st
at
em
en
ts
:
1.
E
ith
er
Jo
hn
w
en
tt
o
th
e
st
or
e,
or
w
e’
re
ou
to
f
eg
gs
.
2.
Jo
e
is
go
in
g
to
le
av
e
ho
m
e
an
d
no
tc
om
e
ba
ck
.
3.
E
ith
er
B
ill
is
at
w
or
k
an
d
Ja
ne
is
n’
t,
or
Ja
ne
is
at
w
or
k
an
d
B
ill
is
n’
t.
05
21
86
12
41
c0
1
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
D
ed
uc
ti
ve
R
ea
so
ni
ng
an
d
L
og
ic
al
C
on
ne
ct
iv
es
11
So
lu
ti
on
s
1.
If
w
e
le
tP
st
an
d
fo
r
th
e
st
at
em
en
t“
Jo
hn
w
en
tt
o
th
e
st
or
e”
an
d
Q
st
an
d
fo
r
“W
e’
re
ou
to
f
eg
gs
,”
th
en
th
is
st
at
em
en
tc
ould
be
re
pr
es
en
te
d
sy
m
bo
lic
al
ly
as
P
∨
Q
.
2.
If
w
e
le
tP
st
an
d
fo
rt
he
st
at
em
en
t“
Jo
e
is
go
in
g
to
le
av
e
ho
m
e”
an
d
Q
st
an
d
fo
r
“J
oe
is
no
tg
oi
ng
to
co
m
e
ba
ck
,”
th
en
w
e
co
ul
d
re
pr
es
en
tt
hi
s
st
at
em
en
t
sy
m
bo
lic
al
ly
as
P
∧
Q
.B
ut
th
is
an
al
ys
is
m
is
se
s
an
im
po
rt
an
tf
ea
tu
re
of
th
e
st
at
em
en
t,
be
ca
us
e
it
do
es
n’
t
in
di
ca
te
th
at
Q
is
a
ne
ga
tiv
e
st
at
em
en
t.
W
e
co
ul
d
ge
ta
be
tte
r
an
al
ys
is
by
le
tti
ng
R
st
an
d
fo
r
th
e
st
at
em
en
t“
Jo
e
is
go
in
g
to
co
m
e
ba
ck
”
an
d
th
en
w
ri
tin
g
th
e
st
at
em
en
t
Q
as
¬R
.P
lu
gg
in
g
th
is
in
to
ou
r
fir
st
an
al
ys
is
of
th
e
or
ig
in
al
st
at
em
en
t,
w
e
ge
t
th
e
im
pr
ov
ed
an
al
ys
is
P
∧
¬
R
.
3.
L
et
B
st
an
d
fo
rt
he
st
at
em
en
t“
B
ill
is
at
w
or
k”
an
d
J
fo
rt
he
st
at
em
en
t“
Ja
ne
is
at
w
or
k.
”
T
he
n
th
e
fir
st
ha
lf
of
th
e
st
at
em
en
t,
“B
ill
is
at
w
or
k
an
d
Ja
ne
is
n’
t,”
ca
n
be
re
pr
es
en
te
d
as
B
∧
¬
J
.
Si
m
ila
rl
y,
th
e
se
co
nd
ha
lf
is
J
∧
¬
B
.
To
re
pr
es
en
tt
he
en
tir
e
st
at
em
en
t,
w
e
m
us
tc
om
bi
ne
th
es
e
tw
o
w
ith
or
,f
or
m
in
g
th
ei
r
di
sj
un
ct
io
n,
so
th
e
so
lu
tio
n
is
(B
∧
¬
J
)∨
(J
∧
¬
B
).
N
ot
ic
e
th
at
in
an
al
yz
in
g
th
e
th
ir
d
st
at
em
en
t
in
th
e
pr
ec
ed
in
g
ex
am
pl
e,
w
e
ad
de
d
pa
re
nt
he
se
s
w
he
n
w
e
fo
rm
ed
th
e
di
sj
un
ct
io
n
of
B
∧
¬
J
an
d
J
∧
¬
B
to
in
di
ca
te
un
am
bi
gu
ou
sl
y
w
hi
ch
st
at
em
en
ts
w
er
e
be
in
g
co
m
bi
ne
d.
T
hi
s
is
lik
e
th
e
us
e
of
pa
re
nt
he
se
s
in
al
ge
br
a,
in
w
hi
ch
,
fo
r
ex
am
pl
e,
th
e
pr
od
uc
t
of
a
+
b
an
d
a
−
b
w
ou
ld
be
w
ri
tte
n
(a
+
b)
·(
a
−
b)
,
w
ith
th
e
pa
re
nt
he
se
s
se
rv
in
g
to
in
di
ca
te
un
am
bi
gu
ou
sl
y
w
hi
ch
qu
an
tit
ie
s
ar
e
to
be
m
ul
tip
lie
d.
A
s
in
al
ge
br
a,
it
is
co
nv
en
ie
nt
in
lo
gi
c
to
om
it
so
m
e
pa
re
nt
he
se
s
to
m
ak
e
ou
r
ex
pr
es
si
on
s
sh
or
te
r
an
d
ea
si
er
to
re
ad
.H
ow
ev
er
,w
e
m
us
ta
gr
ee
on
so
m
e
co
n-
ve
nt
io
ns
ab
ou
t
ho
w
to
re
ad
su
ch
ex
pr
es
si
on
s
so
th
at
th
ey
ar
e
st
ill
un
am
bi
gu
-
ou
s.
O
ne
co
nv
en
tio
n
is
th
at
th
e
sy
m
bo
l
¬
al
w
ay
s
ap
pl
ie
s
on
ly
to
th
e
st
at
e-
m
en
tt
ha
tc
om
es
im
m
ed
ia
te
ly
af
te
ri
t.
Fo
re
xa
m
pl
e,
¬
P
∧
Q
m
ea
ns
(¬
P
)∧
Q
ra
th
er
th
an
¬(
P
∧
Q
).
W
e’
ll
se
e
so
m
e
ot
he
r
co
nv
en
tio
ns
ab
ou
t
pa
re
nt
he
se
s
la
te
r.
E
xa
m
pl
e
1.
1.
3.
W
ha
t
E
ng
lis
h
se
nt
en
ce
s
ar
e
re
pr
es
en
te
d
by
th
e
fo
llo
w
in
g
ex
pr
es
si
on
s?
1.
(¬
S
∧
L
)∨
S,
w
he
re
S
st
an
ds
fo
r“
Jo
hn
is
st
up
id
”
an
d
L
st
an
ds
fo
r“
Jo
hn
is
la
zy
.”
2.
¬S
∧
(L
∨
S)
,w
he
re
S
an
d
L
ha
ve
th
e
sa
m
e
m
ea
ni
ng
s
as
be
fo
re
.
3.
¬(
S
∧
L
)∨
S,
w
ith
S
an
d
L
st
ill
as
be
fo
re
.
05
21
86
12
41
c0
1
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
12
Se
nt
en
ti
al
L
og
ic
So
lu
ti
on
s
1.
E
ith
er
Jo
hn
is
n’
ts
tu
pi
d
an
d
he
is
la
zy
,o
r
he
’s
st
up
id
.
2.
Jo
hn
is
n’
ts
tu
pi
d,
an
d
ei
th
er
he
’s
la
zy
or
he
’s
st
up
id
.N
ot
ic
e
ho
w
th
e
pl
ac
e-
m
en
t
of
th
e
w
or
d
ei
th
er
in
E
ng
lis
h
ch
an
ge
s
ac
co
rd
in
g
to
w
he
re
th
e
pa
re
n-
th
es
es
ar
e.
3.
E
ith
er
Jo
hn
is
n’
t
bo
th
st
up
id
an
d
la
zy
,
or
Jo
hn
is
st
up
id
.
T
he
w
or
d
bo
th
in
E
ng
lis
h
al
so
he
lp
s
di
st
in
gu
is
h
th
e
di
ff
er
en
t
po
ss
ib
le
po
si
tio
ns
of
pa
re
nt
he
se
s.
It
is
im
po
rt
an
t
to
ke
ep
in
m
in
d
th
at
th
e
sy
m
bo
ls
∧,
∨,
an
d
¬
do
n’
t
re
al
ly
co
rr
es
po
nd
to
al
lu
se
s
of
th
e
w
or
ds
an
d,
or
,a
nd
no
t
in
E
ng
lis
h.
Fo
r
ex
am
pl
e,
th
e
sy
m
bo
l
∧
co
ul
d
no
t
be
us
ed
to
re
pr
es
en
t
th
e
us
e
of
th
e
w
or
d
an
d
in
th
e
se
nt
en
ce
“J
oh
n
an
d
B
ill
ar
e
fr
ie
nd
s,
”
be
ca
us
e
in
th
is
se
nt
en
ce
th
e
w
or
d
an
d
is
no
tb
ei
ng
us
ed
to
co
m
bi
ne
tw
o
st
at
em
en
ts
.T
he
sy
m
bo
ls
∧
an
d
∨
ca
n
on
ly
be
us
ed
be
tw
ee
n
tw
o
st
at
em
en
ts
,t
o
fo
rm
th
ei
r
co
nj
un
ct
io
n
or
di
sj
un
ct
io
n,
an
d
th
e
sy
m
bo
l
¬
ca
n
on
ly
be
us
ed
be
fo
re
a
st
at
em
en
t,
to
ne
ga
te
it.
T
hi
s
m
ea
ns
th
at
ce
rt
ai
n
st
ri
ng
s
of
le
tte
rs
an
d
sy
m
bo
ls
ar
e
si
m
pl
y
m
ea
ni
ng
le
ss
.
Fo
r
ex
am
pl
e,
P
¬
∧
Q
,
P
∧/
∨
Q
,
an
d
P
¬Q
ar
e
al
l
“u
ng
ra
m
m
at
ic
al
”
ex
pr
es
si
on
s
in
th
e
la
ng
ua
ge
of
lo
gi
c.
“G
ra
m
m
at
ic
al
”
ex
pr
es
si
on
s,
su
ch
as
th
os
e
in
E
xa
m
pl
es
1.
1.
2
an
d
1.
1.
3,
ar
e
so
m
et
im
es
ca
lle
d
w
el
l-
fo
rm
ed
fo
rm
ul
as
or
ju
st
fo
rm
ul
as
.O
nc
e
ag
ai
n,
it
m
ay
be
he
lp
fu
l
to
th
in
k
of
an
an
al
og
y
w
ith
al
ge
br
a,
in
w
hi
ch
th
e
sy
m
bo
ls
+,
−,
·,
an
d
÷
ca
n
be
us
ed
be
tw
ee
n
tw
o
nu
m
be
rs
,a
s
op
er
at
or
s,
an
d
th
e
sy
m
bo
l
−
ca
n
al
so
be
us
ed
be
fo
re
a
nu
m
be
r,
to
ne
ga
te
it.
T
he
se
ar
e
th
e
on
ly
w
ay
s
th
at
th
es
e
sy
m
bo
ls
ca
n
be
us
ed
in
al
ge
br
a,
so
ex
pr
es
si
on
s
su
ch
as
x
−
÷
y
ar
e
m
ea
ni
ng
le
ss
.
So
m
et
im
es
,w
or
ds
ot
he
r
th
an
an
d,
or
,a
nd
no
ta
re
us
ed
to
ex
pr
es
s
th
e
m
ea
n-
in
gs
re
pr
es
en
te
d
by
∧,
∨,
an
d
¬.
Fo
r
ex
am
pl
e,
co
ns
id
er
th
e
fir
st
st
at
em
en
t
in
E
xa
m
pl
e
1.
1.
3.
A
lth
ou
gh
w
e
ga
ve
th
e
E
ng
lis
h
tr
an
sl
at
io
n
“E
ith
er
Jo
hn
is
n’
t
st
up
id
an
d
he
is
la
zy
,o
r
he
’s
st
up
id
,”
an
al
te
rn
at
iv
e
w
ay
of
co
nv
ey
in
g
th
e
sa
m
e
in
fo
rm
at
io
n
w
ou
ld
be
to
sa
y
“E
ith
er
Jo
hn
is
n’
t
st
up
id
bu
t
he
is
la
zy
,
or
he
’s
st
up
id
.”
O
ft
en
,t
he
w
or
d
bu
t
is
us
ed
in
E
ng
lis
h
to
m
ea
n
an
d,
es
pe
ci
al
ly
w
he
n
th
er
e
is
so
m
e
co
nt
ra
st
or
co
nfl
ic
tb
et
w
ee
n
th
e
st
at
em
en
ts
be
in
g
co
m
bi
ne
d.
Fo
r
a
m
or
e
st
ri
ki
ng
ex
am
pl
e,
im
ag
in
e
a
w
ea
th
er
fo
re
ca
st
er
en
di
ng
hi
s
fo
re
ca
st
w
ith
th
e
st
at
em
en
t
“R
ai
n
an
d
sn
ow
ar
e
th
e
on
ly
tw
o
po
ss
ib
ili
tie
s
fo
r
to
m
or
ro
w
’s
w
ea
th
er
.”
T
hi
s
is
ju
st
a
ro
un
da
bo
ut
w
ay
of
sa
yi
ng
th
at
it
w
ill
ei
th
er
ra
in
or
sn
ow
to
m
or
ro
w
.T
hu
s,
ev
en
th
ou
gh
th
e
fo
re
ca
st
er
ha
s
us
ed
th
e
w
or
d
an
d,
th
e
m
ea
ni
ng
ex
pr
es
se
d
by
hi
s
st
at
em
en
t
is
a
di
sj
un
ct
io
n.
T
he
le
ss
on
of
th
es
e
ex
-
am
pl
es
is
th
at
to
de
te
rm
in
e
th
e
lo
gi
ca
l
fo
rm
of
a
st
at
em
en
t
yo
u
m
us
t
th
in
k
ab
ou
tw
ha
tt
he
st
at
em
en
tm
ea
ns
,r
at
he
rt
ha
n
ju
st
tr
an
sl
at
in
g
w
or
d
by
w
or
d
in
to
sy
m
bo
ls
.
05
21
86
12
41
c0
1
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
D
ed
uc
ti
ve
R
ea
so
ni
ng
an
d
L
og
ic
al
C
on
ne
ct
iv
es
13
So
m
et
im
es
lo
gi
ca
lw
or
ds
ar
e
hi
dd
en
w
ith
in
m
at
he
m
at
ic
al
no
ta
tio
n.
Fo
r
ex
-
am
pl
e,
co
ns
id
er
th
e
st
at
em
en
t
3
≤
π
.
A
lth
ou
gh
it
ap
pe
ar
s
to
be
a
si
m
pl
e
st
at
em
en
t
th
at
co
nt
ai
ns
no
w
or
ds
of
lo
gi
c,
if
yo
u
re
ad
it
ou
t
lo
ud
yo
u
w
ill
he
ar
th
e
w
or
d
or
.
If
w
e
le
t
P
st
an
d
fo
r
th
e
st
at
em
en
t
3
<
π
an
d
Q
fo
r
th
e
st
at
em
en
t
3
=
π
,
th
en
th
est
at
em
en
t
3
≤
π
w
ou
ld
be
w
ri
tte
n
P
∨
Q
.
In
th
is
ex
am
pl
e
th
e
st
at
em
en
ts
re
pr
es
en
te
d
by
th
e
le
tte
rs
P
an
d
Q
ar
e
so
sh
or
t
th
at
it
ha
rd
ly
se
em
s
w
or
th
w
hi
le
to
ab
br
ev
ia
te
th
em
w
ith
si
ng
le
le
tte
rs
.I
n
ca
se
s
lik
e
th
is
w
e
w
ill
so
m
et
im
es
no
tb
ot
he
r
to
re
pl
ac
e
th
e
st
at
em
en
ts
w
ith
le
tte
rs
,s
o
w
e
m
ig
ht
al
so
w
ri
te
th
is
st
at
em
en
ta
s
(3
<
π
)∨
(3
=
π
).
Fo
ra
sl
ig
ht
ly
m
or
e
co
m
pl
ic
at
ed
ex
am
pl
e,
co
ns
id
er
th
e
st
at
em
en
t
3
≤
π
<
4.
T
hi
s
st
at
em
en
t
m
ea
ns
3
≤
π
an
d
π
<
4,
so
on
ce
ag
ai
n
a
w
or
d
of
lo
gi
c
ha
s
be
en
hi
dd
en
in
m
at
he
m
at
ic
al
no
ta
tio
n.
Fi
lli
ng
in
th
e
m
ea
ni
ng
th
at
w
e
ju
st
w
or
ke
d
ou
t
fo
r
3
≤
π
,w
e
ca
n
w
ri
te
th
e
w
ho
le
st
at
em
en
t
as
[(
3
<
π
)∨
(3
=
π
)]
∧
(π
<
4)
.
K
no
w
in
g
th
at
th
e
st
at
em
en
t
ha
s
th
is
lo
gi
ca
l
fo
rm
m
ig
ht
be
im
po
rt
an
t
in
un
de
rs
ta
nd
in
g
a
pi
ec
e
of
m
at
he
m
at
ic
al
re
as
on
in
g
in
vo
lv
in
g
th
is
st
at
em
en
t.
E
xe
rc
is
es
∗ 1
.
A
na
ly
ze
th
e
lo
gi
ca
lf
or
m
s
of
th
e
fo
llo
w
in
g
st
at
em
en
ts
:
(a
)
W
e’
ll
ha
ve
ei
th
er
a
re
ad
in
g
as
si
gn
m
en
to
rh
om
ew
or
k
pr
ob
le
m
s,
bu
tw
e
w
on
’t
ha
ve
bo
th
ho
m
ew
or
k
pr
ob
le
m
s
an
d
a
te
st
.
(b
)
Y
ou
w
on
’t
go
sk
iin
g,
or
yo
u
w
ill
an
d
th
er
e
w
on
’t
be
an
y
sn
ow
.
(c
)
√
7
�≤
2.
2.
A
na
ly
ze
th
e
lo
gi
ca
lf
or
m
s
of
th
e
fo
llo
w
in
g
st
at
em
en
ts
:
(a
)
E
ith
er
Jo
hn
an
d
B
ill
ar
e
bo
th
te
lli
ng
th
e
tr
ut
h,
or
ne
ith
er
of
th
em
is
.
(b
)
I’
ll
ha
ve
ei
th
er
fis
h
or
ch
ic
ke
n,
bu
tI
w
on
’t
ha
ve
bo
th
fis
h
an
d
m
as
he
d
po
ta
to
es
.
(c
)
3
is
a
co
m
m
on
di
vi
so
r
of
6,
9,
an
d
15
.
3.
A
na
ly
ze
th
e
lo
gi
ca
lf
or
m
s
of
th
e
fo
llo
w
in
g
st
at
em
en
ts
:
(a
)
A
lic
e
an
d
B
ob
ar
e
no
tb
ot
h
in
th
e
ro
om
.
(b
)
A
lic
e
an
d
B
ob
ar
e
bo
th
no
ti
n
th
e
ro
om
.
(c
)
E
ith
er
A
lic
e
or
B
ob
is
no
ti
n
th
e
ro
om
.
(d
)
N
ei
th
er
A
lic
e
no
r
B
ob
is
in
th
e
ro
om
.
4.
W
hi
ch
of
th
e
fo
llo
w
in
g
ex
pr
es
si
on
s
ar
e
w
el
l-
fo
rm
ed
fo
rm
ul
as
?
(a
)
¬(
¬
P
∨
¬¬
R
).
(b
)
¬(
P
,
Q
,
∧R
).
(c
)
P
∧
¬
P
.
(d
)
(P
∧
Q
)(
P
∨
R
).
05
21
86
12
41
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1
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
14
Se
nt
en
ti
al
L
og
ic
∗ 5
.
L
et
P
st
an
d
fo
rt
he
st
at
em
en
t“
Iw
ill
bu
y
th
e
pa
nt
s”
an
d
S
fo
rt
he
st
at
em
en
t
“I
w
ill
bu
y
th
e
sh
ir
t.”
W
ha
tE
ng
lis
h
se
nt
en
ce
s
ar
e
re
pr
es
en
te
d
by
th
e
fo
l-
lo
w
in
g
ex
pr
es
si
on
s?
(a
)
¬(
P
∧
¬S
).
(b
)
¬
P
∧
¬S
.
(c
)
¬
P
∨
¬S
.
6.
L
et
S
st
an
d
fo
rt
he
st
at
em
en
t“
St
ev
e
is
ha
pp
y”
an
d
G
fo
r“
G
eo
rg
e
is
ha
pp
y.
”
W
ha
tE
ng
lis
h
se
nt
en
ce
s
ar
e
re
pr
es
en
te
d
by
th
e
fo
llo
w
in
g
ex
pr
es
si
on
s?
(a
)
(S
∨
G
)∧
(¬
S
∨
¬G
).
(b
)
[S
∨
(G
∧
¬S
)]
∨
¬G
.
(c
)
S
∨
[G
∧
(¬
S
∨
¬G
)]
.
7.
Id
en
tif
y
th
e
pr
em
is
es
an
d
co
nc
lu
si
on
s
of
th
e
fo
llo
w
in
g
de
du
ct
iv
e
ar
gu
-
m
en
ts
an
d
an
al
yz
e
th
ei
rl
og
ic
al
fo
rm
s.
D
o
yo
u
th
in
k
th
e
re
as
on
in
g
is
va
lid
?
(A
lth
ou
gh
yo
u
w
ill
ha
ve
on
ly
yo
ur
in
tu
iti
on
to
gu
id
e
yo
u
in
an
sw
er
in
g
th
is
la
st
qu
es
tio
n,
in
th
e
ne
xt
se
ct
io
n
w
e
w
ill
de
ve
lo
p
so
m
e
te
ch
ni
qu
es
fo
r
de
te
rm
in
in
g
th
e
va
lid
ity
of
ar
gu
m
en
ts
.)
(a
)
Ja
ne
an
d
Pe
te
w
on
’t
bo
th
w
in
th
e
m
at
h
pr
iz
e.
Pe
te
w
ill
w
in
ei
th
er
th
e
m
at
h
pr
iz
e
or
th
e
ch
em
is
tr
y
pr
iz
e.
Ja
ne
w
ill
w
in
th
e
m
at
h
pr
iz
e.
T
he
re
fo
re
,P
et
e
w
ill
w
in
th
e
ch
em
is
tr
y
pr
iz
e.
(b
)
T
he
m
ai
n
co
ur
se
w
ill
be
ei
th
er
be
ef
or
fis
h.
T
he
ve
ge
ta
bl
e
w
ill
be
ei
th
er
pe
as
or
co
rn
.W
e
w
ill
no
th
av
e
bo
th
fis
h
as
a
m
ai
n
co
ur
se
an
d
co
rn
as
a
ve
ge
ta
bl
e.
T
he
re
fo
re
,w
e
w
ill
no
th
av
e
bo
th
be
ef
as
a
m
ai
n
co
ur
se
an
d
pe
as
as
a
ve
ge
ta
bl
e.
(c
)
E
ith
er
Jo
hn
or
B
ill
is
te
lli
ng
th
e
tr
ut
h.
E
ith
er
Sa
m
or
B
ill
is
ly
in
g.
T
he
re
fo
re
,e
ith
er
Jo
hn
is
te
lli
ng
th
e
tr
ut
h
or
Sa
m
is
ly
in
g.
(d
)
E
ith
er
sa
le
s
w
ill
go
up
an
d
th
e
bo
ss
w
ill
be
ha
pp
y,
or
ex
pe
ns
es
w
ill
go
up
an
d
th
e
bo
ss
w
on
’t
be
ha
pp
y.
T
he
re
fo
re
,s
al
es
an
d
ex
pe
ns
es
w
ill
no
t
bo
th
go
up
.
1.
2.
T
ru
th
Ta
bl
es
W
e
sa
w
in
Se
ct
io
n
1.
1
th
at
an
ar
gu
m
en
t
is
va
lid
if
th
e
pr
em
is
es
ca
nn
ot
al
l
be
tr
ue
w
ith
ou
tt
he
co
nc
lu
si
on
be
in
g
tr
ue
as
w
el
l.
T
hu
s,
to
un
de
rs
ta
nd
ho
w
w
or
ds
su
ch
as
an
d,
or
,a
nd
no
ta
ff
ec
tt
he
va
lid
ity
of
ar
gu
m
en
ts
,w
e
m
us
ts
ee
ho
w
th
ey
co
nt
ri
bu
te
to
th
e
tr
ut
h
or
fa
ls
ity
of
st
at
em
en
ts
co
nt
ai
ni
ng
th
em
.
W
he
n
w
e
ev
al
ua
te
th
e
tr
ut
h
or
fa
ls
ity
of
a
st
at
em
en
t,
w
e
as
si
gn
to
it
on
e
of
th
e
la
be
ls
tr
ue
or
fa
ls
e,
an
d
th
is
la
be
li
s
ca
lle
d
its
tr
ut
h
va
lu
e.
It
is
cl
ea
rh
ow
th
e
w
or
d
an
d
co
nt
ri
bu
te
s
to
th
e
tr
ut
h
va
lu
e
of
a
st
at
em
en
tc
on
ta
in
in
g
it.
A
st
at
em
en
t
of
th
e
fo
rm
P
∧
Q
ca
n
on
ly
be
tr
ue
if
bo
th
P
an
d
Q
ar
e
tr
ue
;i
f
ei
th
er
P
or
Q
is
fa
ls
e,
th
en
P
∧
Q
w
ill
be
fa
ls
e
to
o.
B
ec
au
se
w
e
ha
ve
as
su
m
ed
th
at
P
an
d
05
21
86
12
41
c0
1
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
Tr
ut
h
Ta
bl
es
15
Fi
gu
re
1
Q
bo
th
st
an
d
fo
r
st
at
em
en
ts
th
at
ar
e
ei
th
er
tr
ue
or
fa
ls
e,
w
e
ca
n
su
m
m
ar
iz
e
al
l
th
e
po
ss
ib
ili
tie
s
w
ith
th
e
ta
bl
e
sh
ow
n
in
Fi
gu
re
1.
T
hi
s
is
ca
lle
d
a
tr
ut
h
ta
bl
e
fo
r
th
e
fo
rm
ul
a
P
∧
Q
.E
ac
h
ro
w
in
th
e
tr
ut
h
ta
bl
e
re
pr
es
en
ts
on
e
of
th
e
fo
ur
po
ss
ib
le
co
m
bi
na
tio
ns
of
tr
ut
h
va
lu
es
fo
r
th
e
st
at
em
en
ts
P
an
d
Q
.
A
lth
ou
gh
th
es
e
fo
ur
po
ss
ib
ili
tie
s
ca
n
ap
pe
ar
in
th
e
ta
bl
e
in
an
y
or
de
r,
it
is
be
st
to
lis
tt
he
m
sy
st
em
at
ic
al
ly
so
w
e
ca
n
be
su
re
th
at
no
po
ss
ib
ili
tie
s
ha
ve
be
en
sk
ip
pe
d.
T
he
tr
ut
h
ta
bl
e
fo
r
¬
P
is
al
so
qu
ite
ea
sy
to
co
ns
tr
uc
t
be
ca
us
e
fo
r
¬
P
to
be
tr
ue
,
P
m
us
tb
e
fa
ls
e.
T
he
ta
bl
e
is
sh
ow
n
in
Fi
gu
re
2.
Fi
gu
re
2
T
he
tr
ut
h
ta
bl
e
fo
r
P
∨
Q
is
a
lit
tle
tr
ic
ki
er
.
T
he
fir
st
th
re
e
lin
es
sh
ou
ld
ce
rt
ai
nl
y
be
fil
le
d
in
as
sh
ow
n
in
Fi
gu
re
3,
bu
t
th
er
e
m
ay
be
so
m
e
qu
es
tio
n
ab
ou
tt
he
la
st
lin
e.
Sh
ou
ld
P
∨
Q
be
tr
ue
or
fa
ls
e
in
th
e
ca
se
in
w
hi
ch
P
an
d
Q
ar
e
bo
th
tr
ue
?
In
ot
he
r
w
or
ds
,d
oe
s
P
∨
Q
m
ea
n
“P
or
Q
,o
r
bo
th
”
or
do
es
it
m
ea
n
“P
or
Q
bu
tn
ot
bo
th
”?
T
he
fir
st
w
ay
of
in
te
rp
re
tin
g
th
e
w
or
d
or
is
ca
lle
d
th
e
in
cl
us
iv
e
or
(b
ec
au
se
it
in
cl
ud
es
th
e
po
ss
ib
ili
ty
of
bo
th
st
at
em
en
ts
be
in
g
tr
ue
),
an
d
th
e
se
co
nd
is
ca
lle
d
th
e
ex
cl
us
iv
e
or
.I
n
m
at
he
m
at
ic
s,
or
al
w
ay
sm
ea
ns
in
cl
us
iv
e
or
,u
nl
es
s
sp
ec
ifi
ed
ot
he
rw
is
e,
so
w
e
w
ill
in
te
rp
re
t∨
as
in
cl
us
iv
e
or
.
W
e
th
er
ef
or
e
co
m
pl
et
e
th
e
tr
uth
ta
bl
e
fo
r
P
∨
Q
as
sh
ow
n
in
Fi
gu
re
4.
Se
e
ex
er
ci
se
3
fo
r
m
or
e
ab
ou
tt
he
ex
cl
us
iv
e
or
.
Fi
gu
re
3
Fi
gu
re
4
U
si
ng
th
e
ru
le
s
su
m
m
ar
iz
ed
in
th
es
e
tr
ut
h
ta
bl
es
,w
e
ca
n
no
w
w
or
k
ou
tt
ru
th
ta
bl
es
fo
r
m
or
e
co
m
pl
ex
fo
rm
ul
as
.
A
ll
w
e
ha
ve
to
do
is
w
or
k
ou
t
th
e
tr
ut
h
va
lu
es
of
th
e
co
m
po
ne
nt
pa
rt
s
of
a
fo
rm
ul
a,
st
ar
tin
g
w
ith
th
e
in
di
vi
du
al
le
tte
rs
an
d
w
or
ki
ng
up
to
m
or
e
co
m
pl
ex
fo
rm
ul
as
a
st
ep
at
a
tim
e.
05
21
86
12
41
c0
1
C
B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
16
Se
nt
en
ti
al
L
og
ic
E
xa
m
pl
e
1.
2.
1.
M
ak
e
a
tr
ut
h
ta
bl
e
fo
r
th
e
fo
rm
ul
a
¬(
P
∨
¬Q
).
So
lu
ti
on
P
Q
¬Q
P
∨
¬Q
¬(
P
∨
¬Q
)
F
F
T
T
F
F
T
F
F
T
T
F
T
T
F
T
T
F
T
F
T
he
fir
st
tw
o
co
lu
m
ns
of
th
is
ta
bl
e
lis
t
th
e
fo
ur
po
ss
ib
le
co
m
bi
na
tio
ns
of
tr
ut
h
va
lu
es
of
P
an
d
Q
.T
he
th
ir
d
co
lu
m
n,
lis
tin
g
tr
ut
h
va
lu
es
fo
r
th
e
fo
rm
ul
a
¬Q
,i
s
fo
un
d
by
si
m
pl
y
ne
ga
tin
g
th
e
tr
ut
h
va
lu
es
fo
r
Q
in
th
e
se
co
nd
co
lu
m
n.
T
he
fo
ur
th
co
lu
m
n,
fo
r
th
e
fo
rm
ul
a
P
∨
¬Q
,i
s
fo
un
d
by
co
m
bi
ni
ng
th
e
tr
ut
h
va
lu
es
fo
r
P
an
d
¬Q
lis
te
d
in
th
e
fir
st
an
d
th
ir
d
co
lu
m
ns
,
ac
co
rd
in
g
to
th
e
tr
ut
h
va
lu
e
ru
le
fo
r∨
su
m
m
ar
iz
ed
in
Fi
gu
re
4.
A
cc
or
di
ng
to
th
is
ru
le
,P
∨
¬Q
w
ill
be
fa
ls
e
on
ly
if
bo
th
P
an
d
¬Q
ar
e
fa
ls
e.
L
oo
ki
ng
in
th
e
fir
st
an
d
th
ir
d
co
lu
m
ns
,w
e
se
e
th
at
th
is
ha
pp
en
s
on
ly
in
ro
w
tw
o
of
th
e
ta
bl
e,
so
th
e
fo
ur
th
co
lu
m
n
co
nt
ai
ns
an
F
in
th
e
se
co
nd
ro
w
an
d
T
’s
in
al
lo
th
er
ro
w
s.
Fi
na
lly
,t
he
tr
ut
h
va
lu
es
fo
r
th
e
fo
rm
ul
a
¬(
P
∨
¬Q
)
ar
e
lis
te
d
in
th
e
fif
th
co
lu
m
n,
w
hi
ch
is
fo
un
d
by
ne
ga
tin
g
th
e
tr
ut
h
va
lu
es
in
th
e
fo
ur
th
co
lu
m
n.
(N
ot
e
th
at
th
es
e
co
lu
m
ns
ha
d
to
be
w
or
ke
d
ou
t
in
or
de
r,
be
ca
us
e
ea
ch
w
as
us
ed
in
co
m
pu
tin
g
th
e
ne
xt
.)
E
xa
m
pl
e
1.
2.
2.
M
ak
e
a
tr
ut
h
ta
bl
e
fo
r
th
e
fo
rm
ul
a
¬(
P
∧
Q
)∨
¬
R
.
So
lu
ti
on
P
Q
R
P
∧
Q
¬(
P
∧
Q
)
¬
R
¬(
P
∧
Q
)∨
¬
R
F
F
F
F
T
T
T
F
F
T
F
T
F
T
F
T
F
F
T
T
T
F
T
T
F
T
F
T
T
F
F
F
T
T
T
T
F
T
F
T
F
T
T
T
F
T
F
T
T
T
T
T
T
F
F
F
N
ot
e
th
at
be
ca
us
e
th
is
fo
rm
ul
a
co
nt
ai
ns
th
re
e
le
tte
rs
,i
t
ta
ke
s
ei
gh
t
lin
es
to
lis
t
al
l
po
ss
ib
le
co
m
bi
na
tio
ns
of
tr
ut
h
va
lu
es
fo
r
th
es
e
le
tte
rs
.
(I
f
a
fo
rm
ul
a
co
nt
ai
ns
n
di
ff
er
en
tl
et
te
rs
,h
ow
m
an
y
lin
es
w
ill
its
tr
ut
h
ta
bl
e
ha
ve
?)
H
er
e’
sa
w
ay
of
m
ak
in
g
tr
ut
h
ta
bl
es
m
or
e
co
m
pa
ct
ly
.I
ns
te
ad
of
us
in
g
se
pa
ra
te
co
lu
m
ns
to
lis
tt
he
tr
ut
h
va
lu
es
fo
r
th
e
co
m
po
ne
nt
pa
rt
s
of
a
fo
rm
ul
a,
ju
st
lis
t
th
os
e
tr
ut
h
va
lu
es
be
lo
w
th
e
co
rr
es
po
nd
in
g
co
nn
ec
tiv
e
sy
m
bo
li
n
th
e
or
ig
in
al
fo
rm
ul
a.
T
hi
s
is
ill
us
tr
at
ed
in
Fi
gu
re
5,
fo
r
th
e
fo
rm
ul
a
fr
om
E
xa
m
pl
e
1.
2.
1.
05
21
86
12
41
c0
1
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B
99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
Tr
ut
h
Ta
bl
es
17
In
th
e
fir
st
st
ep
,w
e
ha
ve
lis
te
d
th
e
tr
ut
h
va
lu
es
fo
r
P
an
d
Q
be
lo
w
th
es
e
le
tte
rs
w
he
re
th
ey
ap
pe
ar
in
th
e
fo
rm
ul
a.
In
st
ep
tw
o,
th
e
tr
ut
h
va
lu
es
fo
r
¬Q
ha
ve
be
en
ad
de
d
un
de
rt
he
¬
sy
m
bo
lf
or
¬Q
.I
n
th
e
th
ir
d
st
ep
,w
e
ha
ve
co
m
bi
ne
d
th
e
tr
ut
h
va
lu
es
fo
r
P
an
d
¬Q
to
ge
tt
he
tr
ut
h
va
lu
es
fo
r
P
∨
¬Q
,w
hi
ch
ar
e
lis
te
d
un
de
rt
he
∨
sy
m
bo
l.
Fi
na
lly
,i
n
th
e
la
st
st
ep
,t
he
se
tr
ut
h
va
lu
es
ar
e
ne
ga
te
d
an
d
lis
te
d
un
de
rt
he
in
iti
al
¬
sy
m
bo
l.
T
he
tr
ut
h
va
lu
es
ad
de
d
in
th
e
la
st
st
ep
gi
ve
th
e
tr
ut
h
va
lu
e
fo
r
th
e
en
tir
e
fo
rm
ul
a,
so
w
e
w
ill
ca
ll
th
e
sy
m
bo
lu
nd
er
w
hi
ch
th
ey
ar
e
lis
te
d
(t
he
fir
st
¬
sy
m
bo
li
n
th
is
ca
se
)
th
e
m
ai
n
co
nn
ec
ti
ve
of
th
e
fo
rm
ul
a.
N
ot
ic
e
th
at
th
e
tr
ut
h
va
lu
es
lis
te
d
un
de
r
th
e
m
ai
n
co
nn
ec
tiv
e
in
th
is
ca
se
ag
re
e
w
ith
th
e
va
lu
es
w
e
fo
un
d
in
E
xa
m
pl
e
1.
2.
1.
Fi
gu
re
5
N
ow
th
at
w
e
kn
ow
ho
w
to
m
ak
e
tr
ut
h
ta
bl
es
fo
r
co
m
pl
ex
fo
rm
ul
as
,
w
e’
re
re
ad
y
to
re
tu
rn
to
th
e
an
al
ys
is
of
th
e
va
lid
ity
of
ar
gu
m
en
ts
.C
on
si
de
r
ag
ai
n
ou
r
fir
st
ex
am
pl
e
of
a
de
du
ct
iv
e
ar
gu
m
en
t:
It
w
ill
ei
th
er
ra
in
or
sn
ow
to
m
or
ro
w
.
It
’s
to
o
w
ar
m
fo
r
sn
ow
.
T
he
re
fo
re
,i
tw
ill
ra
in
.
A
s
w
e
ha
ve
se
en
,i
f
w
e
le
t
P
st
an
d
fo
r
th
e
st
at
em
en
t
“I
t
w
ill
ra
in
to
m
or
ro
w
”
an
d
Q
fo
r
th
e
st
at
em
en
t
“I
t
w
ill
sn
ow
to
m
or
ro
w
,”
th
en
w
e
ca
n
re
pr
es
en
t
th
e
ar
gu
m
en
ts
ym
bo
lic
al
ly
as
fo
llo
w
s:
P
∨
Q
¬Q ∴
P
(T
he
sy
m
bo
l
∴
m
ea
ns
th
er
ef
or
e.
)
W
e
ca
n
no
w
se
e
ho
w
tr
ut
h
ta
bl
es
ca
n
be
us
ed
to
ve
ri
fy
th
e
va
lid
ity
of
th
is
ar
gu
m
en
t.
Fi
gu
re
6
sh
ow
s
a
tr
ut
h
ta
bl
e
fo
r
bo
th
pr
em
is
es
an
d
th
e
co
nc
lu
si
on
of
th
e
ar
gu
m
en
t.
R
ec
al
l
th
at
w
e
de
ci
de
d
to
ca
ll
an
ar
gu
m
en
t
va
lid
if
th
e
05
21
86
12
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99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
18
Se
nt
en
ti
al
L
og
ic
pr
em
is
es
ca
nn
ot
al
lb
e
tr
ue
w
ith
ou
tt
he
co
nc
lu
si
on
be
in
g
tr
ue
as
w
el
l.
L
oo
ki
ng
at
Fi
gu
re
6
w
e
se
e
th
at
th
e
on
ly
ro
w
of
th
e
ta
bl
e
in
w
hi
ch
bo
th
pr
em
is
es
co
m
e
ou
tt
ru
e
is
ro
w
th
re
e,
an
d
in
th
is
ro
w
th
e
co
nc
lu
si
on
is
al
so
tr
ue
.T
hu
s,
th
e
tr
ut
h
ta
bl
e
co
nfi
rm
s
th
at
if
th
e
pr
em
is
es
ar
e
al
lt
ru
e,
th
e
co
nc
lu
si
on
m
us
ta
ls
o
be
tr
ue
,
so
th
e
ar
gu
m
en
ti
s
va
lid
.
Fi
gu
re
6
E
xa
m
pl
e
1.
2.
3.
D
et
er
m
in
e
w
he
th
er
th
e
fo
llo
w
in
g
ar
gu
m
en
ts
ar
e
va
lid
.
1.
E
ith
er
Jo
hn
is
n’
ts
tu
pi
d
an
d
he
is
la
zy
,o
r
he
’s
st
up
id
.
Jo
hn
is
st
up
id
.
T
he
re
fo
re
,J
oh
n
is
n’
tl
az
y.
2.
T
he
bu
tle
r
an
d
th
e
co
ok
ar
e
no
tb
ot
h
in
no
ce
nt
.
E
ith
er
th
e
bu
tle
r
is
ly
in
g
or
th
e
co
ok
is
in
no
ce
nt
.
T
he
re
fo
re
,t
he
bu
tle
r
is
ei
th
er
ly
in
g
or
gu
ilt
y.
So
lu
ti
on
s
1.
A
s
in
E
xa
m
pl
e
1.
1.
3,
w
e
le
tS
st
an
d
fo
r
th
e
st
at
em
en
t“
Jo
hn
is
st
up
id
”
an
d
L
st
an
d
fo
r
“J
oh
n
is
la
zy
.”
T
he
n
th
e
ar
gu
m
en
th
as
th
e
fo
rm
:
(¬
S
∧
L
)∨
S
S ∴
¬
L
N
ow
w
e
m
ak
e
a
tr
ut
h
ta
bl
e
fo
r
bo
th
pr
em
is
es
an
d
th
e
co
nc
lu
si
on
.(
Y
ou
sh
ou
ld
w
or
k
ou
tt
he
in
te
rm
ed
ia
te
st
ep
s
in
de
ri
vi
ng
co
lu
m
n
th
re
e
of
th
is
ta
bl
e
to
co
nfi
rm
th
at
it
is
co
rr
ec
t.)
Pr
em
is
es
C
on
cl
us
io
n
S
L
(¬
S
∧
L
)∨
S
S
¬L
F
F
F
F
T
F
T
T
F
F
T
F
T
T
T
T
T
T
T
F
B
ot
h
pr
em
is
es
ar
e
tr
ue
in
lin
es
th
re
e
an
d
fo
ur
of
th
is
ta
bl
e.
T
he
co
nc
lu
si
on
is
al
so
tr
ue
in
lin
e
th
re
e,
bu
t
it
is
fa
ls
e
in
lin
e
fo
ur
.T
hu
s,
it
is
po
ss
ib
le
fo
r
05
21
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12
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6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
8612
4
1
C
ha
r
C
ou
nt
=
0
Tr
ut
h
Ta
bl
es
19
bo
th
pr
em
is
es
to
be
tr
ue
an
d
th
e
co
nc
lu
si
on
fa
ls
e,
so
th
e
ar
gu
m
en
ti
s
in
va
lid
.
In
fa
ct
,t
he
ta
bl
e
sh
ow
s
us
ex
ac
tly
w
hy
th
e
ar
gu
m
en
ti
s
in
va
lid
.T
he
pr
ob
le
m
oc
cu
rs
in
th
e
fo
ur
th
lin
e
of
th
e
ta
bl
e,
in
w
hi
ch
S
an
d
L
ar
e
bo
th
tr
ue
–
in
ot
he
r
w
or
ds
,J
oh
n
is
bo
th
st
up
id
an
d
la
zy
.T
hu
s,
if
Jo
hn
is
bo
th
st
up
id
an
d
la
zy
,
th
en
bo
th
pr
em
is
es
w
ill
be
tr
ue
bu
tt
he
co
nc
lu
si
on
w
ill
be
fa
ls
e,
so
it
w
ou
ld
be
a
m
is
ta
ke
to
in
fe
r
th
at
th
e
co
nc
lu
si
on
m
us
tb
e
tr
ue
fr
om
th
e
as
su
m
pt
io
n
th
at
th
e
pr
em
is
es
ar
e
tr
ue
.
2.
L
et
B
st
an
d
fo
r
th
e
st
at
em
en
t“
T
he
bu
tle
r
is
in
no
ce
nt
,”
C
fo
r
th
e
st
at
em
en
t
“T
he
co
ok
is
in
no
ce
nt
,”
an
d
L
fo
r
th
e
st
at
em
en
t“
T
he
bu
tle
r
is
ly
in
g.
”
T
he
n
th
e
ar
gu
m
en
th
as
th
e
fo
rm
:
¬(
B
∧
C
)
L
∨
C
∴
L
∨
¬
B
H
er
e
is
th
e
tr
ut
h
ta
bl
e
fo
r
th
e
pr
em
is
es
an
d
co
nc
lu
si
on
:
Pr
em
is
es
C
on
cl
us
io
n
B
C
L
¬(
B
∧
C
)
L
∨
C
L
∨
¬
B
F
F
F
T
F
T
F
F
T
T
T
T
F
T
F
T
T
T
F
T
T
T
T
T
T
F
F
T
F
F
T
F
T
T
T
T
T
T
F
F
T
F
T
T
T
F
T
T
T
he
pr
em
is
es
ar
e
bo
th
tr
ue
on
ly
in
lin
es
tw
o,
th
re
e,
fo
ur
,
an
d
si
x,
an
d
in
ea
ch
of
th
es
e
ca
se
s
th
e
co
nc
lu
si
on
is
tr
ue
as
w
el
l.
T
he
re
fo
re
,t
he
ar
gu
m
en
t
is
va
lid
.
If
yo
u
ex
pe
ct
ed
th
e
fir
st
ar
gu
m
en
ti
n
E
xa
m
pl
e
1.
2.
3
to
tu
rn
ou
tt
o
be
va
lid
,
it’
s
pr
ob
ab
ly
be
ca
us
e
th
e
fir
st
pr
em
is
e
co
nf
us
ed
yo
u.
It
’s
a
ra
th
er
co
m
pl
ic
at
ed
st
at
em
en
t,
w
hi
ch
w
e
re
pr
es
en
te
d
sy
m
bo
lic
al
ly
w
ith
th
e
fo
rm
ul
a
(¬
S
∧
L
)∨
S.
A
cc
or
di
ng
to
ou
r
tr
ut
h
ta
bl
e,
th
is
fo
rm
ul
a
is
fa
ls
e
if
S
an
d
L
ar
e
bo
th
fa
ls
e,
an
d
tr
ue
ot
he
rw
is
e.
B
ut
no
tic
e
th
at
th
is
is
ex
ac
tly
th
e
sa
m
e
as
th
e
tr
ut
h
ta
bl
e
fo
r
th
e
si
m
pl
er
fo
rm
ul
a
L
∨
S!
B
ec
au
se
of
th
is
,w
e
sa
y
th
at
th
e
fo
rm
ul
as
(¬
S
∧
L
)∨
S
an
d
L
∨
S
ar
e
eq
ui
va
le
nt
.
E
qu
iv
al
en
t
fo
rm
ul
as
al
w
ay
s
ha
ve
th
e
sa
m
e
tr
ut
h
va
lu
e
no
m
at
te
r
w
ha
t
st
at
em
en
ts
th
e
le
tte
rs
in
th
em
st
an
d
fo
r
an
d
no
m
at
te
r
w
ha
tt
he
tr
ut
h
va
lu
es
of
th
os
e
st
at
em
en
ts
ar
e.
T
he
eq
ui
va
le
nc
e
of
th
e
pr
em
is
e
(¬
S
∧
L
)∨
S
an
d
th
e
si
m
pl
er
fo
rm
ul
a
L
∨
S
m
ay
he
lp
yo
u
un
de
rs
ta
nd
w
hy
05
21
86
12
41
c0
1
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99
6/
V
el
le
m
an
O
ct
ob
er
19
,2
00
5
23
:4
6
0
52
1
86
12
4
1
C
ha
r
C
ou
nt
=
0
20
Se
nt
en
ti
al
L
og
ic
th
e
ar
gu
m
en
t
is
in
va
lid
.T
ra
ns
la
tin
g
th
e
fo
rm
ul
a
L
∨
S
ba
ck
in
to
E
ng
lis
h,
w
e
se
e
th
at
th
e
fir
st
pr
em
is
e
co
ul
d
ha
ve
be
en
st
at
ed
m
or
e
si
m
pl
y
as
“J
oh
n
is
ei
th
er
la
zy
or
st
up
id
(o
r
bo
th
).”
B
ut
fr
om
th
is
pr
em
is
e
an
d
th
e
se
co
nd
pr
em
is
e
(t
ha
t
Jo
hn
is
st
up
id
),
it
cl
ea
rl
y
do
es
n’
tf
ol
lo
w
th
at
he
’s
no
tl
az
y,
be
ca
us
e
he
m
ig
ht
be
bo
th
st
up
id
an
d
la
zy
.
E
xa
m
pl
e
1.
2.
4.
W
hi
ch
of
th
es
e
fo
rm
ul
as
ar
e
eq
ui
va
le
nt
?
¬(
P
∧
Q
),
¬
P
∧
¬Q
,
¬
P
∨
¬Q
.
So
lu
ti
on
H
er
e’
s
a
tr
ut
h
ta
bl
e
fo
r
al
lt
hr
ee
st
at
em
en
ts
.(
Y
ou
sh
ou
ld
ch
ec
k
it
yo
ur
se
lf
!)
P
Q
¬(
P
∧
Q
)
¬
P
∧
¬Q
¬
P
∨
¬Q
F
F
T
T
T
F
T
T
F
T
T
F
T
F
T
T
T
F
F
F
T
he
th
ir
d
an
d
fif
th
co
lu
m
ns
in
th
is
ta
bl
e
ar
e
id
en
tic
al
,b
ut
th
ey
ar
e
di
ff
er
en
t
fr
om
th
e
fo
ur
th
co
lu
m
n.
T
he
re
fo
re
,
th
e
fo
rm
ul
as
¬(
P
∧
Q
)
an
d
¬
P
∨
¬Q
ar
e
eq
ui
va
le
nt
,b
ut
ne
ith
er
is
eq
ui
va
le
nt
to
th
e
fo
rm
ul
a
¬
P
∧
¬Q
.T
hi
s
sh
ou
ld
m
ak
e
se
ns
e
if
yo
u
th
in
k
ab
ou
tw
ha
ta
ll
th
e
sy
m
bo
ls
m
ea
n.
Fo
re
xa
m
pl
e,
su
pp
os
e
P
st
an
ds
fo
r
th
e
st
at
em
en
t
“T
he
Y
an
ke
es
w
on
la
st
ni
gh
t”
an
d
Q
st
an
ds
fo
r
“T
he
R
ed
So
x
w
on
la
st
ni
gh
t.”
T
he
n
¬(
P
∧
Q
)
w
ou
ld
m
ea
n
“T
he
Y
an
ke
es
an
d
th
e
R
ed
So
x
di
d
no
t
bo
th
w
in
la
st
ni
gh
t,”
an
d
¬
P
∨
¬Q
w
ou
ld
m
ea
n
“E
ith
er
th
e
Y
an
ke
es
or
th
e
R
ed
So
x
lo
st
la
st
ni
gh
t”
;
th
es
e
st
at
em
en
ts
cl
ea
rl
y
co
nv
ey
th
e
sa
m
e
in
fo
rm
at
io
n.
O
n
th
e
ot
he
rh
an
d,
¬
P
∧
¬Q
w
ou
ld
m
ea
n
“T
he
Y
an
ke
es
an
d
th
e
R
ed
So
x
bo
th
lo
st
la
st
ni
gh
t,”
w
hi
ch
is
an
en
tir
el
y
di
ff
er
en
t
st
at
em
en
t.
Y
ou
ca
n
ch
ec
k
fo
ry
ou
rs
el
fb
y
m
ak
in
g
a
tr
ut
h
ta
bl
e
th
at
th
e
fo
rm
ul
a
¬
P
∧
¬Q
fr
om
E
xa
m
pl
e
1.
2.
4
is
eq
ui
va
le
nt
to
th
e
fo
rm
ul
a
¬(
P
∨
Q
).
(T
o
se
e
th
at
th
is
eq
ui
va
le
nc
e
m
ak
es
se
ns
e,
no
tic
e
th
at
th
e
st
at
em
en
ts
“B
ot
h
th
e
Y
an
ke
es
an
d
th
e
R
ed
So
x
lo
st
la
st
ni
gh
t”
an
d
“N
ei
th
er
th
e
Y
an
ke
es
no
r
th
e
R
ed
So
x
w
on
la
st
ni
gh
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hi
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e
an
d
th
e
on
e
di
sc
ov
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ed
in
E
xa
m
pl
e
1.
2.
4
ar
e
ca
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d
D
eM
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la
w
s.
In
an
al
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d
th
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st
at
em
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th
at
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cu
r
in
th
em
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is
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lp
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l
to
be
fa
m
ili
ar
w
ith
a
nu
m
be
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of
eq
ui
va
le
nc
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th
at
co
m
e
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of
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er
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e
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ui
va
le
nc
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in
th
e
fo
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w
in
g
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t
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ak
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ak
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as
in
to
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ng
lis
h,
as
w
e
di
d
in
E
xa
m
pl
e
1.
2.
4.
05
21
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=
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Tr
ut
h
Ta
bl
es
21
D
eM
or
ga
n’
s
la
w
s
¬(
P
∧
Q
)
is
eq
ui
va
le
nt
to
¬
P
∨
¬Q
.
¬(
P
∨
Q
)
is
eq
ui
va
le
nt
to
¬
P
∧
¬Q
.
C
om
m
ut
at
iv
e
la
w
s
P
∧
Q
is
eq
ui
va
le
nt
to
Q
∧
P
.
P
∨
Q
is
eq
ui
va
le
nt
to
Q
∨
P
.
A
ss
oc
ia
ti
ve
la
w
s
P
∧
(Q
∧
R
)
is
eq
ui
va
le
nt
to
(P
∧
Q
)∧
R
.
P
∨
(Q
∨
R
)
is
eq
ui
va
le
nt
to
(P
∨
Q
)∨
R
.
Id
em
po
te
nt
la
w
s
P
∧
P
is
eq
ui
va
le
nt
to
P
.
P
∨
P
is
eq
ui
va
le
nt
to
P
.
D
is
tr
ib
ut
iv
e
la
w
s
P
∧
(Q
∨
R
)
is
eq
ui
va
le
nt
to
(P
∧
Q
)∨
(P
∧
R
).
P
∨
(Q
∧
R
)
is
eq
ui
va
le
nt
to
(P
∨
Q
)∧
(P
∨
R
).
A
bs
or
pt
io
n
la
w
s
P
∨
(P
∧
Q
)
is
eq
ui
va
le
nt
to
P
.
P
∧
(P
∨
Q
)
is
eq
ui
va
le
nt
to
P
.
D
ou
bl
e
N
eg
at
io
n
la
w
¬¬
P
is
eq
ui
va
le
nt
to
P.
N
ot
ic
e
th
at
be
ca
us
e
of
th
e
as
so
ci
at
iv
e
la
w
s
w
e
ca
n
le
av
e
ou
tp
ar
en
th
es
es
in
fo
rm
ul
as
of
th
e
fo
rm
s
P
∧
Q
∧
R
an
d
P
∨
Q
∨
R
w
ith
ou
t
w
or
ry
in
g
th
at
th
e
re
su
lti
ng
fo
rm
ul
a
w
ill
be
am
bi
gu
ou
s,
be
ca
us
e
th
e
tw
o
po
ss
ib
le
w
ay
s
of
fil
lin
g
in
th
e
pa
re
nt
he
se
s
le
ad
to
eq
ui
va
le
nt
fo
rm
ul
as
.
M
an
y
of
th
e
eq
ui
va
le
nc
es
in
th
e
lis
ts
ho
ul
d
re
m
in
d
yo
u
of
si
m
ila
r
ru
le
s
in
-
vo
lv
in
g
+,
·,a
nd
−
in
al
ge
br
a.
A
s
in
al
ge
br
a,
th
es
e
ru
le
s
ca
n
be
ap
pl
ie
d
to
m
or
e
co
m
pl
ex
fo
rm
ul
as
,a
nd
th
ey
ca
n
be
co
m
bi
ne
d
to
w
or
k
ou
t
m
or
e
co
m
pl
ic
at
ed
eq
ui
va
le
nc
es
.A
ny
of
th
e
le
tte
rs
in
th
es
e
eq
ui
va
le
nc
es
ca
n
be
re
pl
ac
ed
by
m
or
e
co
m
pl
ic
at
ed
fo
rm
ul
as
,a
nd
th
e
re
su
lti
ng
eq