You should be really careful when doing this. The fact that two
compositions
(T1 @ {#a -> #b} @ {#x -> #y})
and
(T1 @ {#x -> #y} @ {#a -> #b})
are semantically the same does not mean that the way in which
they are expressed in class/trait definition is not important
of some importance for programmer.
That's pretty much like indentation - code which is properly
indented with nice long variable names is could be semantically
equivalent to the code which is total mess in a single line
but to programmers the difference is huge - at least for me.
If you really need semantic/structural equivalence, you'd better
flatten protocols of both compositions and compare these. This would
handle correctly even this case:
TX = (T1 + T3)
(T1 + T2 + T3) <==> (T2 + TX)
Best, Jan
On 22/01/14 09:33, Camille Teruel wrote:
On 22 janv. 2014, at 10:13, Martin Dias <[email protected]
<mailto:[email protected]>> wrote:
On Wed, Jan 22, 2014 at 10:05 AM, Martin Dias <[email protected]
<mailto:[email protected]>> wrote:
should
(T1 @ {#a -> #b} @ {#x -> #y}) = (T1 @ {#x -> #y} @ {#a -> #b})
?
two more:
(T1 - {#a. #b}) = (T1 - {#b. #a})
(T1 - {#a} - {#b}) = (T1 - {#b} - {#a})
I would say yes for each case.
