This is a file in the archives of the Stanford Encyclopedia of Philosophy. |
(1) For all agents i1, i2, , imHence,N,
![]()
Ki1Ki2 Kim(A)
Proof.
Note first that
( 2 ) |
![]() i 1 ![]() |
K i 1 ( |
![]() i 2 ![]() |
K i 2 ( . . . ( |
![]() i m - 1 ![]() |
K i m - 1 ( |
![]() i m ![]() |
K i m( A ) ) ) ) ) |
= |
![]() i 1 ![]() |
K i 1 ( |
![]() i 2 ![]() |
K i 2 ( . . . ( |
![]() i m - 1 ![]() |
K i m - 1 (K 1N ( A ) ) ) ) ) |
= |
![]() i 1 ![]() |
K i 1 ( |
![]() i 2 ![]() |
K i 2 . . . ( |
![]() i m - 2 ![]() |
K i m - 2 ( K 2N ( A ) ) ) ) |
= . . . |
= |
![]() i 1 ![]() |
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 1N
K i 1 ( ![]()
i 2N
K i 2 ( . . . ( ![]()
i m - 1N
K i m - 1 ( ![]()
i mN
K i m ( A ) ) ) ) )
so by ( 2 ), if ( 1 ) is satisfied then
K mN ( A ).
Peter Vanderschraaf peterv@MAIL1.ANDREW.CMU.EDU |