X-Git-Url: https://git.stderr.nl/gitweb?p=matthijs%2Fmaster-project%2Freport.git;a=blobdiff_plain;f=Chapters%2FNormalization.tex;h=d564d231405a00ab7f31622ef525b15f987c2218;hp=b058b63a2a3debae92f4fe435fe1f31799555c40;hb=d92aa4307ca45f07c6ae50056e08ffc874839756;hpb=8638abef4686335ede35793b43f3e25930822544 diff --git a/Chapters/Normalization.tex b/Chapters/Normalization.tex index b058b63..d564d23 100644 --- a/Chapters/Normalization.tex +++ b/Chapters/Normalization.tex @@ -1590,7 +1590,7 @@ \subsubsection[sec:transformation:caseremoval]{Case removal} This transform removes any case expression with a single alternative and - only wild binders.\refdef{wild binder} + only wild binders.\refdef{wild binders} These "useless" case expressions are usually leftovers from case simplification on extractor case (see the previous example). @@ -2240,11 +2240,13 @@ developed in the final part of the research, leaving no more time for verifying these properties. In fact, it is likely that the current transformation system still violates some of these - properties in some cases and should be improved (or extra conditions - on the input hardware descriptions should be formulated). + properties in some cases (see + \in{section}[sec:normalization:non-determinism] and + \in{section}[sec:normalization:stateproblems]) and should be improved (or + extra conditions on the input hardware descriptions should be formulated). This is most likely the case with the completeness and determinism - properties, perhaps als the termination property. The soundness + properties, perhaps also the termination property. The soundness property probably holds, since it is easier to manually verify (each transformation can be reviewed separately). @@ -2441,7 +2443,7 @@ each node in the normal set is also in the intended normal set. Reasoning about our intended normal set is easier, since we know how to generate it from its definition. \refdef{intended normal - form definition}. + form definition} Fortunately, we can also prove the complement (which is equivalent, since $A \subseteq B \Leftrightarrow \overline{B}