Re: [isabelle] Simplifying addition and subtraction of multisets

Hi Manuel,

I also think that you need a simproc, because the nesting can be arbitrarily deep. Since multisets are similar to natural numbers in terms of their algebraic properties, I'd suggest that you look at the simprocs for natural numbers. I guess that you can generalise cancel_diff_conv in src/HOL/Tools/Nat_Arith.ML appropriately to multisets.

Hope this helps,

On 11/04/16 13:55, Manuel Eberl wrote:
Well, that works for some cases, but not for all, e.g:

lemma "(A + B + C + D ) - (C :: 'a multiset) = A + B + D"

I don't think this is going to work without a simproc. The arithmetic procedures do things
like that; maybe they can be adapted for this, or perhaps it's just a matter of the right

I don't know who is the expert on these procedures.


On 08/04/16 14:24, Julian Nagele wrote:
Hello Manuel,

simplifying with subset_mset.diff_add_assoc works for me:

lemma "{#a, b, c, d#} - {#b#} = {#a, c, d#}"
   by (simp add: subset_mset.diff_add_assoc)

or, more generally

   fixes A B C :: "'a multiset"
   shows "(A + C + B) - C = A + B"
   by (simp add: subset_mset.diff_add_assoc ac_simps)

Hope that helps,

Manuel Eberl writes:


I have terms like "{#a,b,c,d#} - {#b#}", which desugars to "(single a +
single b + single c + single d) - single b". This is obviously equal to

However, the simplifier fails to prove this and I was not able to find a
setup of existing simplification rules to solve it.

I ended up proving the following rule, which works in my particular case:

lemma multiset_Diff_single_normalize:
    fixes a c assumes "a â c"
    shows   "({#a#} + B) - {#c#} = {#a#} + (B - {#c#})"

This, combined with add_ac, does the trick. (but only because I can, in
this particular instance, decide whether two elements are equal, i.e. I
know that b != c and b != d, even though the simplification of
"{#a,b,c,d#} - {#b#}" would be sound even if that were not the case)

Is there some existing simproc that can be set up to do this automatically?



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