If E
F, then as we observed earlier, Ki(E)
Ki(F), so
| This is a file in the archives of the Stanford Encyclopedia of Philosophy. |
K*N(E) and
E
F, then
K*N(F).
Proof.
If E
F, then as we observed earlier, Ki(E)
Ki(F), so
K 1N ( E ) = 
iN
K i( E ) =
iKi(F) = K1N(F)
If we now set
E
=
KnN(E) and
F
=
KnN(F), then by the argument just given we have
Kn+1N(E) = K1N(E
so we have mth level mutual knowledge for every
n 
1.
|
Hence if
|
n = 1 |
KnN(E) then
|
n = 1 |
KnN(F). 
|
| Peter Vanderschraaf peterv@cyrus.andrew.cmu.edu |