It looks like you're asking if substitution (@) is commutative. In the general case it's not: T1 @ {#x -> #y} @ {#y -> #z} == T1 but T1 @ {#y -> #z} @ {#x -> #y} is not. I think Jan's suggestion of flattening the traits and then comparing them for semantic/structural equivalence is the right way to avoid going crazy! frank On 22 January 2014 09:05, Martin Dias <tinchodias@gmail.com> wrote:
should
(T1 @ {#a -> #b} @ {#x -> #y}) = (T1 @ {#x -> #y} @ {#a -> #b})
?
On Tue, Jan 21, 2014 at 6:39 PM, Martin Dias <tinchodias@gmail.com> wrote:
On Tue, Jan 21, 2014 at 6:02 PM, Damien Cassou <damien.cassou@gmail.com> wrote:
On Tue, Jan 21, 2014 at 5:29 PM, Martin Dias <tinchodias@gmail.com> wrote:
Do experts on traits agree? I can open an issue and propose a slice.
you have to pay attention that A + B + C = B + A + C = C + A + B.
ok, I will propose a slice