Open DavePearce opened 3 years ago
If you want to allow overlapping lvals, I think it should generate a proof obligation:
assert i != j;
or perhaps more generally:
assert i == j ==> rhs1 == rhs2 (0 == 1) in this case.
Hey,
Yeah, so that second obligation is interesting. I'll have to have a think, but yeah definitely one of these two makes sense.
Does this make sense:
What is the semantics of this? Dafny disallows this.