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| -rw-r--r-- | main.tex | 2 |
1 files changed, 1 insertions, 1 deletions
@@ -222,7 +222,7 @@ based on what they evaluate to. For case I think that would end up being a three \JP{I suppose this would essentially be an ``interpreter'' style semantics but we can prove equivalence pretty easily.} -\YH{To add to that, I used to have both in the Coq code, but commented the recursive definition out, and now only have the inductive definition, which is basically what I copy pasted here.} \JW{Fair enough. Whatever you think ends up being the easiest to read and understand, really. There's something to be said for staying close to the Coq definitions anyway.} \YH{I have added more rules, we can always switch from one to the other now. One more thing I noticed though is that recursive definitions will need an \texttt{option} type.} \JW{Oh, then my suggestion of `stmntrun f s0 (Vseq st1 st2) = stmntrun f (stmntrun f s0 st1) st2' is a bit ill-typed then, unless the second parameter becomes option-typed too. Maybe the inference rules are better overall then.} +\YH{To add to that, I used to have both in the Coq code, but commented the recursive definition out, and now only have the inductive definition, which is basically what I copy pasted here.} \JW{Fair enough. Whatever you think ends up being the easiest to read and understand, really. There's something to be said for staying close to the Coq definitions anyway.} \YH{I have added more rules, we can always switch from one to the other now. One more thing I noticed though is that recursive definitions will need an \texttt{option} type.} \JW{Oh, then my suggestion of `stmntrun f s0 (Vseq st1 st2) = stmntrun f (stmntrun f s0 st1) st2' is a bit ill-typed then, unless the second parameter becomes option-typed too. Maybe the inference rules are better overall then.} \YH{Ah yes, I actually didn't even notice that, it would need the do notation, just like the implementation in Coq, so it may be easier to just use the inference rules.} \begin{align} \text{srun}\ f\ s\ \mathtt{Vskip} &= \mathtt{Some}\ s\\ |