At 13:46 +0100 1999/06/03, Peter Hancock wrote:
>Just regard log_x ... as alternative
>notation for \ x -> ...
>
>log_x (a * b) = log_x a + log_x b
>log_x 1 = 0
>log_x x = 1
>
>log_x (a ^ b) = (log_x a) * b , x not free in b .
These are interesting relations in some sense, even though it is hard to
immediately see what to do with them.
What one can do though is to take say the primitive set S = { I, +, *, ^ }
plus the obvious relations R they satisfy (like associativity, and some
other such relations). Then is the question is, is (S; R) equivalent to the
set of all lambda expressions? -- The difference is that when using the
primitive set S only, we subsume the relations in the algebra of lambda
expressions, but here we try to express these relations explicitly.
Hans Aberg
* Email: Hans Aberg <mailto:[EMAIL PROTECTED]>
* Home Page: <http://www.matematik.su.se/~haberg/>
* AMS member listing: <http://www.ams.org/cml/>
