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Equivalencesinpredicatelogic

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Unformatted text preview: true.

 2.  But
if
P(x)
denotes

“x
>
0,” then ∃!x
P(x)
is
false.
   The
uniqueness
quantifier
is
not
really
needed
as
the
restriction
 that
there
is
a
unique
x
such
that
P(x)
can
be
expressed
as:


 






























∃x
(P(x)
∧∀y
(P(y)
→ y =x))
 Thinking
about
Quan*fiers
   When
the

domain
of
discourse
is
finite,
we
can
think
of
 quantification
as
looping
through
the
elements
of
the
domain.
   To
evaluate
∀x
P(x)
loop
through
all
x
in
the
domain.

   If
at
every
step
P(x)
is
true,
then
∀x
P(x)
is
true.

   If
at
a
step
P(x)
is
false,
then
∀x
P(x)
is
false
and
the
loop
 terminates.

   To
evaluate
∃x
P(x)
loop
through
all
x
in
the
domain.

   If

at
some
step,
P(x)
is
true,
then
∃x
P(x)
is
true
and
the
loop
 terminates.

   If
the
loop
ends
without
finding
an
x
for
which
P(x)
is
true,
then
∃x
 P(x)
is
false.
   Even
if
the
domains
are
infinite,
we
can
still
think
of
the
 quantifiers
this
fashion,
but
the
loops
will
not
terminate
in
 some
cases.
 Proper*es
of
Quan*fiers
   The
truth
value
of
∃x P(x)

and
∀ x P(x) depend on both the propositional function P(x) and on the domain U.   Examples: If
U
is
the

positive
integers
and
P(x)
is
the
statement










 “x
<
2”,
then
∃x P(x)


is
true,
but
∀ x P(x) is false. 2.  If
Uis
the
negative
integers
and
P(x)
is
the
statement










 “x
<
2”,
then
both
∃x P(x)

and

∀ x P(x) are true. 3.  If
U
consists
of
3,
4,
and
5,

and
P(x)
is
the
statement










 “x
>
2”,
then

both
∃x P(x)


and
∀ x P(x) are true....
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This document was uploaded on 03/06/2014 for the course MATH 320 at CSU Northridge.

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