[ The Types Forum, http://lists.seas.upenn.edu/mailman/listinfo/types-list ]
Dear Ahmed,
I'm not sure I fully understand what you are asking, if you're looking
for a total order on the set of simply-typed lambda-terms such that t
--> t' implies t < t', then you can do as follows.
In the simply-typed lambda-calculus, every term strongly normalizes, so
you may define an integer #t which is equal to the length of the longest
reduction of t to its normal form. This induces a partition on the set
of simply-typed lambda-terms in countably many sets, call them S_n, such
that S_n contains all terms t s.t. #t = n. Each S_n is of course
countable, so you may pick an injection beta_n : S_n ---> N, which
induces a total ordering within S_n. Observe now that t --> t' implies
#t' < #t, which means that no two terms in the same S_n beta-reduce to
one another. Therefore, the map
t |---> (#t, beta_n(t)).
induces a total order on the simply-typed lambda-terms isomorphic to
omega^2 with the desired property.
Best wishes,
Damiano
On 06/01/2019 12:54, Ahmed Bhayat wrote:
[ The Types Forum, http://lists.seas.upenn.edu/mailman/listinfo/types-list ]
Dear all,
Would welcome information on the following:
Does there exist a reduction ordering, total on ground terms of the
simply-typed lambda calculus, that is compatible with beta-reduction?
That is, does there exist a reduction order ‘>’, total on ground terms of STT,
such that for all terms t1 and t2 such that t1 beta-reduces to t2, we have t1 > t2?
Alternatively, is there a proof that no such ordering can exist?
Failing either of the above, I would be interested in any partial results or
possibly relevant results.
Thanks
Ahmed Bhayat
[email protected]
PhD candidate at the University of Manchester
