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Proof.
If E
F, then as we observed earlier, K_{i}(E)
K_{i}(F), so
K^{ 1}_{N} ( E ) =
^{i N}K_{ i}( E ) =
^{i N}K_{i}(F) = K^{1}_{N}(F)
If we now set E = K^{n}_{N}(E) and F = K^{n}_{N}(F), then by the argument just given we have
K^{n+1}_{N}(E) = K^{1}_{N}(E) K^{1}_{N}(F) = K^{n+1}_{N}(F)
so we have m^{th} level mutual knowledge for every n 1.
Hence if |
^{n = 1} |
K^{n}_{N}(E) then |
^{n = 1} |
K^{n}_{N}(F). |
First published: August 27, 2001
Content last modified: August 27, 2001