/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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NO
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Outermost Termination of the given OTRS could be disproven:

(0) OTRS
(1) OutermostNonTerminationProof [COMPLETE, 2766 ms]
(2) NO


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(0)
Obligation:
Term rewrite system R:
The TRS R consists of the following rules:

   sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
   sel(0, cons(X, Z)) -> X
   first(0, Z) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
   from(X) -> cons(X, n__from(s(X)))
   sel1(s(X), cons(Y, Z)) -> sel1(X, activate(Z))
   sel1(0, cons(X, Z)) -> quote(X)
   first1(0, Z) -> nil1
   first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, activate(Z)))
   quote(n__0) -> 01
   quote1(n__cons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
   quote1(n__nil) -> nil1
   quote(n__s(X)) -> s1(quote(activate(X)))
   quote(n__sel(X, Z)) -> sel1(activate(X), activate(Z))
   quote1(n__first(X, Z)) -> first1(activate(X), activate(Z))
   unquote(01) -> 0
   unquote(s1(X)) -> s(unquote(X))
   unquote1(nil1) -> nil
   unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
   fcons(X, Z) -> cons(X, Z)
   first(X1, X2) -> n__first(X1, X2)
   from(X) -> n__from(X)
   0 -> n__0
   cons(X1, X2) -> n__cons(X1, X2)
   nil -> n__nil
   s(X) -> n__s(X)
   sel(X1, X2) -> n__sel(X1, X2)
   activate(n__first(X1, X2)) -> first(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__0) -> 0
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(n__nil) -> nil
   activate(n__s(X)) -> s(X)
   activate(n__sel(X1, X2)) -> sel(X1, X2)
   activate(X) -> X



Outermost Strategy.

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(1) OutermostNonTerminationProof (COMPLETE)
Term rewrite system R:
The TRS R consists of the following rules:

   sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
   sel(0, cons(X, Z)) -> X
   first(0, Z) -> nil
   first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
   from(X) -> cons(X, n__from(s(X)))
   sel1(s(X), cons(Y, Z)) -> sel1(X, activate(Z))
   sel1(0, cons(X, Z)) -> quote(X)
   first1(0, Z) -> nil1
   first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, activate(Z)))
   quote(n__0) -> 01
   quote1(n__cons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
   quote1(n__nil) -> nil1
   quote(n__s(X)) -> s1(quote(activate(X)))
   quote(n__sel(X, Z)) -> sel1(activate(X), activate(Z))
   quote1(n__first(X, Z)) -> first1(activate(X), activate(Z))
   unquote(01) -> 0
   unquote(s1(X)) -> s(unquote(X))
   unquote1(nil1) -> nil
   unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z))
   fcons(X, Z) -> cons(X, Z)
   first(X1, X2) -> n__first(X1, X2)
   from(X) -> n__from(X)
   0 -> n__0
   cons(X1, X2) -> n__cons(X1, X2)
   nil -> n__nil
   s(X) -> n__s(X)
   sel(X1, X2) -> n__sel(X1, X2)
   activate(n__first(X1, X2)) -> first(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__0) -> 0
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(n__nil) -> nil
   activate(n__s(X)) -> s(X)
   activate(n__sel(X1, X2)) -> sel(X1, X2)
   activate(X) -> X



Outermost Strategy.

---------- Loop: ----------

quote1(activate(n__from(X'))) -> quote1(from(X')) with rule activate(n__from(X'')) -> from(X'') at position [0] and matcher [X'' / X']

quote1(from(X')) -> quote1(cons(X', n__from(s(X')))) with rule from(X) -> cons(X, n__from(s(X))) at position [0] and matcher [X / X']

quote1(cons(X', n__from(s(X')))) -> quote1(n__cons(X', n__from(s(X')))) with rule cons(X1, X2) -> n__cons(X1, X2) at position [0] and matcher [X1 / X', X2 / n__from(s(X'))]

quote1(n__cons(X', n__from(s(X')))) -> cons1(quote(activate(X')), quote1(activate(n__from(s(X'))))) with rule quote1(n__cons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z))) at position [] and matcher [X / X', Z / n__from(s(X'))]

Now an instance of the first term with Matcher [X' / s(X')] occurs in the last term at position [1].

Context: cons1(quote(activate(X')), [])

We used [THIEMANN_LOOPS_UNDER_STRATEGIES] to show that this Loop is an Outermost-Loop.
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(2)
NO