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 <[email protected]> wrote:
> should
>
> (T1 @ {#a -> #b} @ {#x -> #y}) = (T1 @ {#x -> #y} @ {#a -> #b})
>
> ?
>
>
>
> On Tue, Jan 21, 2014 at 6:39 PM, Martin Dias <[email protected]> wrote:
>>
>>
>>
>>
>> On Tue, Jan 21, 2014 at 6:02 PM, Damien Cassou <[email protected]>
>> wrote:
>>>
>>> On Tue, Jan 21, 2014 at 5:29 PM, Martin Dias <[email protected]>
>>> 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
>
>
