THE TALMUD RULE AND THE SECUREMENT OF ... - ResearchGate

for which EE. Now, let us denote by s 1 the minimum positive
integer for which E< ( 1 2 )s 1+r 1 +1 +
r 1
j=0 ( 1 2 )j · z. Since R satisfies selfduality
and restricted path independence,
k=1
k=1
R(N,E,(z,z)) = (b 1 ,b 1 )+R(N,E 1 , (e 1 ,e 1 )),
where b 1 = 1 · z · r 1
2 j=0 ( 1 2 )j , E 1 = E − z · r 1
j=0 ( 1 2 )j ,ande 1 =( 1 2 )s 1+r 1 +1 · z.
Since 1 ≤ min{r 1 ,s 1 },thene 1 ≤ 1z. Notice that now E 4 1 >e 1 and we can
apply the same argument. After a finite number of iterations, say K, we
would have
K K
R(N,E,(z,z)) = ( b k , b k )+R(N,E K , (e k ,e k )),
k=1
k=1
where e k ≤ K
1
· z. Asaresult,R
4
i (N,E K , (e k ,e k )) ≤ 1
· z, fori =1, 2.
4K
If we take limits, when K →∞,then
∞ ∞
R(N,E,(z, z)) = ( b k , b k ),
k=1
which implies R 1 (N,E,(z,z)) = R 2 (N,E,(z, z)).
The main result of the paper is the following:
6
k=1