Closed garrigue closed 5 years ago
Another possible syntax, but with a change of semantics, would be
\rsum_(x = d #> s) F x
meaning x
equal to d
on s
(i.e. remove the implicit complement).
I tried the other syntax, but while it nicely matches the sub_vec
syntax (t # V
), it doesn't work well with things like:
\rsum_(x = d #> ~:('V m0 :\ n0)) F x
Change
into
meaning
x
equal tod
on the complement ofs
.