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m ∈N
n m C
ʴ n , h ( lfp (
ʴ n , h ))
C
by Lemma 4.6 (2)
n ∈N C
ʴ n , h ( lfp (
ʴ n , h ))
=
C
n ∈N
ʴ n , h ) .
=
lfp (
C
This completes the proof of ( 4.20 ).
(0, 1] and ʔ is a subdistribution in a pLTS
Lemma 4.8
Suppose ʴ
S , ʩ ˄ ,
.
If
T , ʘ ,
is a resolution of ʔ , then we have
V
ʴ , h ( ʘ )
≤ V
ʴ , max ( ʔ ) for any
reward vector h.
Proof
Let f : T S be the resolving function associated with the resolution
T , ʘ ,
, we show by induction on n that
ʴ , h , n
ʴ , h , n ( t ) for any t T.
V
max ( f ( t ))
≥ V
(4.21)
ˉ
−ₒ
The base case n
=
0 is trivial. We consider the inductive step. If t
ʘ , then
ˉ
−ₒ
ˉ
ʴ , h , n
ʴ , h , n ( t ). Now suppose that t
f ( t )
f ( ʘ ), thus
V
max ( f ( t ))
=
h ( ˉ )
= V
−ₒ
and
˄
−ₒ
˄
−ₒ
ˉ
t
ʘ . Then, f ( t )
−ₒ
and f ( t )
f ( ʘ ). We can infer that
˄
−ₒ
ʴ , h ,( n + 1)
max
ʴ , h , n
V
( f ( t ))
=
ʴ
·
max
{V
max ( ʔ )
|
f ( t )
ʔ
}
ʴ , h , n
ʴ · V
max ( f ( ʘ ))
· s S f ( ʘ )( s )
ʴ , h , n
=
ʴ
· V
max ( s )
· t T ʘ ( t )
ʴ , h , n
max ( f ( t ))
=
ʴ
· V
· t T ʘ ( t )
ʴ , h , n ( t )
ʴ
· V
by induction
ʴ , h , n ( ʘ )
= ʴ · V
ʴ , h ,( n
+
1) ( t ) .
= V
So we have proved ( 4.21 ), from which it follows that
ʴ , h
max ( f ( t ))
ʴ , h ( t ) for any t
V
≥ V
T.
(4.22)
Therefore, we have that
ʴ , h
max ( ʔ )
ʴ , h
max ( f ( ʘ ))
V
= V
s S f ( ʘ )( s )
ʴ , max ( s )
=
· V
t T ʘ ( t )
ʴ , h
=
· V
max ( f ( t ))
t T ʘ ( t )
ʴ , h ( t )
· V
( 4.22 )
ʴ , h ( ʘ ) .
= V
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