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x
=
ʱ
Hence
= 1 if and only if both of the conjuncts are 1. The second conjunct
is 1, i.e.,
ʲ
⃒
∈
x
(1
)=1;
ʲ∈
dom(
ʱ
)
if and only if for each
ʲ
ʲ
∈
dom(
ʱ
), 1
⃒
∈
x
= 1 i.e.,
y
=
ʲ
1
⃒
(
x
(
y
)
∧
)=1;
y
∈
dom(
x
)
if and only if for each
ʲ
∈
dom(
ʱ
)thereexists
y
∈
dom(
x
) such that
x
(
y
)
∈
y
=
ʲ
= 1; if and only if for each
ʲ
{
1
,
1
/
2
}
and
∈
dom(
ʱ
)thereexists
y
∈
dom(
x
) such that
y
is
ʲ
-like (by the induction hypothesis) and
x
(
y
)
∈{
1
,
1
/
2
}
.
Again, since the first conjunct is 1, we have,
(
x
(
y
)
⃒
y
∈
ʱ
)=1;
y
∈
dom(
x
)
if and only if for each
y
∈
dom(
x
), (
x
(
y
)
⃒
y
∈
ʱ
) = 1, i.e.,
y
=
ʲ
(
x
(
y
)
⃒
dom(
ʱ
)
)=1;
ʲ
∈
if and only if for each
y
∈
dom(
x
), if
y
is not
ʲ
-like for any
ʲ
∈
ʱ
then by
induction hypothesis it can be derived that
x
(
y
)=0.
Hence combining the above results we get
= 1 if and only if
x
is
ʱ
-like and hence by the (meta-) induction the proof is done.
x
=
ʱ
V
(PS
3
)
and ʱ
Lemma 8.
For any x
∈
∈
ORD
,
x
∈
ʱ
=1
if and only if x is
ʲ-like for some ʲ
∈
ʱ.
Proof.
Using Lemma 7, the following three statements are equivalent:
= 1 if and only if
u∈
dom(
ʱ
)
1.
=1;
2. there exists
ʲ ∈
dom(
ʱ
) such that
x
=
ʲ
=1;and
3.
x
is
ʲ
-like for some
ʲ
x
∈
ʱ
x
=
u
∈
ʱ
.
ORD, there are many
ʱ
-like
names in
V
(PS
3
)
in addition to
ʱ
. Of course, we would desire that
ʱ
-like names
are equal and that for
ʲ<ʱ
,
ʲ
-like names are elements of
ʱ
-like names (in the
formal sense of
V
(
A
)
):
It is clear from the definition that for any
ʱ
∈
V
(PS
3
)
be ʱ-like for some ʱ
V
(PS
3
)
,
Theorem 9.
Let x
∈
∈
ORD
. For any y
∈
x
=
y
=1
if and only if y is ʱ-like.
