N,
Ki1Ki2 … Kim(A)
| This is a file in the archives of the Stanford Encyclopedia of Philosophy. |
KmN(A)
iff
(1) For all agents i1, i2, … , imHence,
K*N(A)
iff (1) is the case for each m
1.
Proof.
Note first that
| ( 2 ) |
i 1  N
|
K i 1 ( |
i 2 N
|
K i 2 ( . . . ( |
i m - 1 N
|
K i m - 1 ( |
i m N
|
K i m( A ) ) ) ) ) |
| = |
i 1 N
|
K i 1 ( |
i 2 N
|
K i 2 ( . . . ( |
i m - 1 N
|
K i m - 1 (K 1N ( A ) ) ) ) ) |
| = |
i 1 N
|
K i 1 ( |
i 2 N
|
K i 2 . . . ( |
i m - 2 N
|
K i m - 2 ( K 2N ( A ) ) ) ) |
| = . . . |
| = |
i 1 N
|
K i 1( K m - 1N ( A ) ) |
| = | K mN ( A ) |
By ( 2 ), K mN ( A ) 
K i 1K i 2 . . . K i m( A )
for i 1, i 2, . . ., i m N
so if K mN ( A ) then condition ( 1 ) is satisfied. Condition ( 1 ) is equivalent to
i 1K i 1 (
i 2K i 2 ( . . . (
i m - 1K i m - 1 (
i mK i m ( A ) ) ) ) )
so by ( 2 ), if ( 1 ) is satisfied then K mN ( A ).
| Peter Vanderschraaf peterv@cyrus.andrew.cmu.edu |