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Notice that the relation
→
is
idempotent
, that is, it holds that (
→◦→
)=
→
. In particular,
we have that
→◦
e
(
M
)=
e
(
M
)
,
(20.16)
e
(
M
)
◦→
=
e
(
M
)
.
(20.17)
Thus, if
S
1
,
S
2
,
T
are instances such that (
S
1
,
S
2
)
∈→
and (
S
2
,
T
)
∈
e
(
M
),then(
S
1
,
T
)
∈
e
(
M
). Hence, if (
S
1
,
S
2
)
∈→
, then it holds that S
OL
e
(
M
)
(
S
2
)
⊆
S
OL
e
(
M
)
(
S
1
). We use this
property in the proof.
(1) We first show that
M
admits a maximum extended recovery if and only if
e
(
M
) admits
a maximum recovery.
(
M
be a maximum recovery of
e
(
M
is also a
⇐
)Let
M
). We show next that
M
is a recovery of
e
(
maximum extended recovery of
M
.Since
M
),wehavethat
◦M
for every instance
S
of R
1
. Moreover, from
(20.17)
we have that
(
S
,
S
)
∈
e
(
M
)
◦M
=
e
(
◦→◦M
and, thus, (
S
,
S
)
◦→◦M
for every in
e
(
M
)
M
)
∈
e
(
M
)
stance
S
of R
1
. Thus, given that (
S
,
S
)
∈→
for every instance
S
of R
1
, we obtain that
◦→◦M
◦→
M
) for every instance
S
of R
1
, which implies
(
S
,
S
)
∈
M
M
◦
e
(
)
=
e
(
)
e
(
M
is an extended recovery of
M
that
.
M
be an extended recovery of
M
)
Now, let
M
. Then we have that (
S
,
S
)
∈
e
(
M
)
◦
e
(
M
) is a recovery of
e
(
for every instance
S
of R
1
, and, hence, we have that
e
(
). Recall
that
M
is a maximum recovery of
e
(
M
) and, hence, we have that
e
(
M
)
◦M
⊆
e
(
M
)
◦
e
(
M
M
), which implies that
e
(
◦M
◦→⊆
e
(
M
)
M
)
M
)
◦
e
(
◦→
. Therefore, given that
M
)
M
) by
(20.17)
,wehavethat
e
(
e
(
M
)=
e
(
M
)
◦→
and
e
(
◦→
=
e
(
M
)
◦→
◦M
◦→⊆
e
(
M
), which implies that
e
(
M
)
M
). Thus,
M
)
◦
e
(
M
)
◦
e
(
⊆
e
(
M
)
◦
e
(
M
is an extended recovery of
we have shown that
M
, and that for every other extended
M
of
M
)
M
), which implies that
M
recovery
M
, it holds that
e
(
M
)
◦
e
(
⊆
e
(
M
)
◦
e
(
is a maximum extended recovery of
M
.
M
be a maximum extended recovery of
M
) is a
(
⇒
)Let
M
. Next we show that
e
(
maximum recovery of
e
(
M
).
M
is an extended recovery of
M
)
Given that
M
,wehavethat(
S
,
S
)
∈
e
(
M
)
◦
e
(
M
) is a recovery of
e
(
for every instance
S
of R
1
, which implies that
e
(
M
).Thus,by
M
) is a maximum recovery of
e
(
Proposition 20.21
, to prove that
e
(
M
), it is enough to
M
), since this fact
show that S
OL
e
(
M
)
(
S
2
)
⊆
S
OL
e
(
M
)
(
S
1
) for every (
S
1
,
S
2
)
∈
e
(
M
)
◦
e
(
M
)
M
). To prove that
implies that
e
(
M
)
◦
e
(
◦
e
(
M
)
⊆
e
(
M
).Let(
S
1
,
S
2
)
∈
e
(
M
)
◦
e
(
from R
2
to R
1
:
S
OL
e
(
M
)
(
S
2
)
⊆
S
OL
e
(
M
)
(
S
1
), we make use of the following mapping
M
=
{
(
T
,
S
)

S
is an instance of R
1
and (
S
1
,
T
)
∈
e
(
M
)
}∪
{
M
(
T
,
S
)

(
S
1
,
T
)
∈
e
(
M
) and S
OL
e
(
M
)
(
S
)
⊆
S
OL
e
(
M
)
(
S
1
)
}
.
is an extended recovery of
We show first that
M
M
, that is, we show that for every
). First, assume that S
OL
e
(
M
)
(
S
)
instance
S
of R
1
, it holds that (
S
,
S
)
∈
e
(
M
)
◦
e
(
M
⊆
S
OL
e
(
M
)
(
S
1
), and consider an arbitrary instance
T
such that (
S
,
T
)
∈
e
(
M
). Notice that
(
S
1
,
T
)
S
OL
e
(
M
)
(
S
1
). Thus, we have that (
T
,
S
)
and,
∈
e
(
M
) since S
OL
e
(
M
)
(
S
)
⊆
∈M