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[Maude-help] single-identifier in full maude qualifications?


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  • From: mp463634 AT mail.inf.tu-dresden.de
  • To: maude-help AT maude.cs.uiuc.edu
  • Subject: [Maude-help] single-identifier in full maude qualifications?
  • Date: Wed, 28 Jul 2004 17:06:23 +0200 (MEST)
  • Importance: Normal
  • List-archive: <http://maude.cs.uiuc.edu/pipermail/maude-help>
  • List-id: Maude help list <maude-help.maude.cs.uiuc.edu>

Hello,

I am playing around with FullMaude 2.1 and parameterized modules, thinking
of the very simple graph theory with a module for finite paths over a
graph

think of the very simple example code
**********************
(fth graph is
sorts N E .
ops s t : E -> N .
endfth)

(fmod path(G :: graph) is
sort P(G) .
subsort
G@E
< P(G) .
op _;_ : P(G) P(G) ~> P(G) [assoc] .

ops s t : P(G) ->
G@N
.
cmb g:P`(G`) ; h:P`(G`) : P(G) if t(g:P`(G`)) = s(h:P`(G`)) .

var g h : P(G) .

ceq s(g ; h) = s(g) if t(g) = s(h) .
ceq t(g ; h) = t(h) if t(g) = s(h) .
endfm)
**************************

it can directly be executed

while
cmb g:P`(G`) ; h:P`(G`) : P(G) if t(g:P`(G`)) = s(h:P`(G`)) .
may be cumbersome, it can be done, however
ceq s(g ; h) = s(g) if t(g) = s(h) .
cannot stand, it should and in some cases has to be replaced by
ceq s(g ; h) = s(g) if (g ; h) : P(G) .
or anything like this, however i seem unable to get a correct syntax for
this... usually the error goes like: could not find sort "P`(G`)"
but instantiating P(G) to P`(G`) gives parse errors

matching equations like
ceq s(f) = s(g) if g ; h := f .
work but are not always applicable

looking for help, martin


*****************************************
for testing:

(fmod nc is
including graph .

ops A B : -> N .
ops z y : -> E .

eq s(z) = A .
eq t(z) = B .
eq s(y) = B .
eq t(y) = B .
endfm)

(view gnc from graph to nc is
sort N to N .
sort E to E .
endv)





  • [Maude-help] single-identifier in full maude qualifications?, mp463634, 07/28/2004

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