This is my impression too but it is late.
Especially since traits are designed to avoid to take care about structure of
the trait “nesting”.
We do not care that a method come from T1 or T1.T2
Stef
> 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
>>
>>
>
