[isabelle-dev] Why do cancellation simprocs call Lin_Arith.simproc
Tobias Nipkow
nipkow at in.tum.de
Sat Oct 29 16:55:24 CEST 2011
> One set of simprocs will cancel factors from inequalities, rewriting
> terms like "x * z < y * z" to either "x < y" or "y < x", depending on
> whether it can prove "0 < z" or "z < 0". What I don't understand is
> the method they use to try to prove "0 < z" or "z < 0": Instead of
> recursively calling the full simplifier (as the simplifier would do
> when applying a conditional rewrite rule) it just calls the linear
> arithmetic simproc directly. (The code for this is in the function
> sign_conv in HOL/Tools/numeral_simprocs.ML, introduced in rev.
> 57753e0ec1d4.)
I was probably afraid of an excessive number of recursive simplifier
calls in the case of long products. But that is not a killer argument.
Just try it.
Tobias
> This necessitates some awkward/redundant setup of simp rules: For
> example, at type real, we have the cancellation simprocs, which match
> goals like "x * z < y * z" or "z * x < z * y". We also have some
> cancellation simp rules like "0 < z ==> (z * x < z * y) = (x < y)". It
> seems like the simprocs should make the simp rule redundant, but they
> don't: Terms like "real (fact n) * x < real (fact n) * y" can be
> rewritten by the simp rule, but the simproc fails because it cannot
> solve the side-condition "0 < real (fact n)".
>
> Is there any reason why the cancellation simprocs *shouldn't* call the
> full simplifier recursively?
>
> - Brian
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