/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be disproven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 10 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) TransformationProof [EQUIVALENT, 0 ms]
(6) QDP
(7) TransformationProof [EQUIVALENT, 0 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) TransformationProof [EQUIVALENT, 0 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 0 ms]
(14) QDP
(15) TransformationProof [EQUIVALENT, 0 ms]
(16) QDP
(17) NonTerminationLoopProof [COMPLETE, 2 ms]
(18) NO


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

   from(X) -> cons(X, n__from(s(X)))
   length(n__nil) -> 0
   length(n__cons(X, Y)) -> s(length1(activate(Y)))
   length1(X) -> length(activate(X))
   from(X) -> n__from(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(X) -> X

Q is empty.

----------------------------------------

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FROM(X) -> CONS(X, n__from(s(X)))
   LENGTH(n__cons(X, Y)) -> LENGTH1(activate(Y))
   LENGTH(n__cons(X, Y)) -> ACTIVATE(Y)
   LENGTH1(X) -> LENGTH(activate(X))
   LENGTH1(X) -> ACTIVATE(X)
   ACTIVATE(n__from(X)) -> FROM(X)
   ACTIVATE(n__nil) -> NIL
   ACTIVATE(n__cons(X1, X2)) -> CONS(X1, X2)

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(s(X)))
   length(n__nil) -> 0
   length(n__cons(X, Y)) -> s(length1(activate(Y)))
   length1(X) -> length(activate(X))
   from(X) -> n__from(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 6 less nodes.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH(n__cons(X, Y)) -> LENGTH1(activate(Y))
   LENGTH1(X) -> LENGTH(activate(X))

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(s(X)))
   length(n__nil) -> 0
   length(n__cons(X, Y)) -> s(length1(activate(Y)))
   length1(X) -> length(activate(X))
   from(X) -> n__from(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule LENGTH1(X) -> LENGTH(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

   (LENGTH1(n__from(x0)) -> LENGTH(from(x0)),LENGTH1(n__from(x0)) -> LENGTH(from(x0)))
   (LENGTH1(n__nil) -> LENGTH(nil),LENGTH1(n__nil) -> LENGTH(nil))
   (LENGTH1(n__cons(x0, x1)) -> LENGTH(cons(x0, x1)),LENGTH1(n__cons(x0, x1)) -> LENGTH(cons(x0, x1)))
   (LENGTH1(x0) -> LENGTH(x0),LENGTH1(x0) -> LENGTH(x0))


----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH(n__cons(X, Y)) -> LENGTH1(activate(Y))
   LENGTH1(n__from(x0)) -> LENGTH(from(x0))
   LENGTH1(n__nil) -> LENGTH(nil)
   LENGTH1(n__cons(x0, x1)) -> LENGTH(cons(x0, x1))
   LENGTH1(x0) -> LENGTH(x0)

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(s(X)))
   length(n__nil) -> 0
   length(n__cons(X, Y)) -> s(length1(activate(Y)))
   length1(X) -> length(activate(X))
   from(X) -> n__from(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(7) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule LENGTH1(n__from(x0)) -> LENGTH(from(x0)) at position [0] we obtained the following new rules [LPAR04]:

   (LENGTH1(n__from(x0)) -> LENGTH(cons(x0, n__from(s(x0)))),LENGTH1(n__from(x0)) -> LENGTH(cons(x0, n__from(s(x0)))))
   (LENGTH1(n__from(x0)) -> LENGTH(n__from(x0)),LENGTH1(n__from(x0)) -> LENGTH(n__from(x0)))


----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH(n__cons(X, Y)) -> LENGTH1(activate(Y))
   LENGTH1(n__nil) -> LENGTH(nil)
   LENGTH1(n__cons(x0, x1)) -> LENGTH(cons(x0, x1))
   LENGTH1(x0) -> LENGTH(x0)
   LENGTH1(n__from(x0)) -> LENGTH(cons(x0, n__from(s(x0))))
   LENGTH1(n__from(x0)) -> LENGTH(n__from(x0))

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(s(X)))
   length(n__nil) -> 0
   length(n__cons(X, Y)) -> s(length1(activate(Y)))
   length1(X) -> length(activate(X))
   from(X) -> n__from(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH1(n__nil) -> LENGTH(nil)
   LENGTH(n__cons(X, Y)) -> LENGTH1(activate(Y))
   LENGTH1(n__cons(x0, x1)) -> LENGTH(cons(x0, x1))
   LENGTH1(x0) -> LENGTH(x0)
   LENGTH1(n__from(x0)) -> LENGTH(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(s(X)))
   length(n__nil) -> 0
   length(n__cons(X, Y)) -> s(length1(activate(Y)))
   length1(X) -> length(activate(X))
   from(X) -> n__from(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(11) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule LENGTH1(n__nil) -> LENGTH(nil) at position [0] we obtained the following new rules [LPAR04]:

   (LENGTH1(n__nil) -> LENGTH(n__nil),LENGTH1(n__nil) -> LENGTH(n__nil))


----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH(n__cons(X, Y)) -> LENGTH1(activate(Y))
   LENGTH1(n__cons(x0, x1)) -> LENGTH(cons(x0, x1))
   LENGTH1(x0) -> LENGTH(x0)
   LENGTH1(n__from(x0)) -> LENGTH(cons(x0, n__from(s(x0))))
   LENGTH1(n__nil) -> LENGTH(n__nil)

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(s(X)))
   length(n__nil) -> 0
   length(n__cons(X, Y)) -> s(length1(activate(Y)))
   length1(X) -> length(activate(X))
   from(X) -> n__from(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(13) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH1(n__cons(x0, x1)) -> LENGTH(cons(x0, x1))
   LENGTH(n__cons(X, Y)) -> LENGTH1(activate(Y))
   LENGTH1(x0) -> LENGTH(x0)
   LENGTH1(n__from(x0)) -> LENGTH(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(s(X)))
   length(n__nil) -> 0
   length(n__cons(X, Y)) -> s(length1(activate(Y)))
   length1(X) -> length(activate(X))
   from(X) -> n__from(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(15) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule LENGTH1(n__cons(x0, x1)) -> LENGTH(cons(x0, x1)) at position [0] we obtained the following new rules [LPAR04]:

   (LENGTH1(n__cons(x0, x1)) -> LENGTH(n__cons(x0, x1)),LENGTH1(n__cons(x0, x1)) -> LENGTH(n__cons(x0, x1)))


----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH(n__cons(X, Y)) -> LENGTH1(activate(Y))
   LENGTH1(x0) -> LENGTH(x0)
   LENGTH1(n__from(x0)) -> LENGTH(cons(x0, n__from(s(x0))))
   LENGTH1(n__cons(x0, x1)) -> LENGTH(n__cons(x0, x1))

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(s(X)))
   length(n__nil) -> 0
   length(n__cons(X, Y)) -> s(length1(activate(Y)))
   length1(X) -> length(activate(X))
   from(X) -> n__from(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(X1, X2)
   activate(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(17) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = LENGTH1(activate(n__from(x0))) evaluates to  t =LENGTH1(activate(n__from(s(x0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [x0 / s(x0)]
* Semiunifier: [ ]

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Rewriting sequence

LENGTH1(activate(n__from(x0))) -> LENGTH1(n__from(x0))
with rule activate(X) -> X at position [0] and matcher [X / n__from(x0)]

LENGTH1(n__from(x0)) -> LENGTH(cons(x0, n__from(s(x0))))
with rule LENGTH1(n__from(x0')) -> LENGTH(cons(x0', n__from(s(x0')))) at position [] and matcher [x0' / x0]

LENGTH(cons(x0, n__from(s(x0)))) -> LENGTH(n__cons(x0, n__from(s(x0))))
with rule cons(X1, X2) -> n__cons(X1, X2) at position [0] and matcher [X1 / x0, X2 / n__from(s(x0))]

LENGTH(n__cons(x0, n__from(s(x0)))) -> LENGTH1(activate(n__from(s(x0))))
with rule LENGTH(n__cons(X, Y)) -> LENGTH1(activate(Y))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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(18)
NO