/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 4 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) TransformationProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) TransformationProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) DependencyGraphProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPOrderProof [EQUIVALENT, 35 ms]
        (18) QDP
        (19) NonTerminationLoopProof [COMPLETE, 20 ms]
        (20) 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(n__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)
   s(X) -> n__s(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(activate(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(n__s(X)))
   LENGTH(n__cons(X, Y)) -> S(length1(activate(Y)))
   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(activate(X))
   ACTIVATE(n__from(X)) -> ACTIVATE(X)
   ACTIVATE(n__s(X)) -> S(activate(X))
   ACTIVATE(n__s(X)) -> ACTIVATE(X)
   ACTIVATE(n__nil) -> NIL
   ACTIVATE(n__cons(X1, X2)) -> CONS(activate(X1), X2)
   ACTIVATE(n__cons(X1, X2)) -> ACTIVATE(X1)

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(n__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)
   s(X) -> n__s(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(activate(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 2 SCCs with 8 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVATE(n__s(X)) -> ACTIVATE(X)
   ACTIVATE(n__from(X)) -> ACTIVATE(X)
   ACTIVATE(n__cons(X1, X2)) -> ACTIVATE(X1)

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(n__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)
   s(X) -> n__s(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
   activate(X) -> X

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

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVATE(n__s(X)) -> ACTIVATE(X)
   ACTIVATE(n__from(X)) -> ACTIVATE(X)
   ACTIVATE(n__cons(X1, X2)) -> ACTIVATE(X1)

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

(8) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*ACTIVATE(n__s(X)) -> ACTIVATE(X)
The graph contains the following edges 1 > 1


*ACTIVATE(n__from(X)) -> ACTIVATE(X)
The graph contains the following edges 1 > 1


*ACTIVATE(n__cons(X1, X2)) -> ACTIVATE(X1)
The graph contains the following edges 1 > 1


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(9)
YES

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

(10)
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(n__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)
   s(X) -> n__s(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(activate(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(X) -> LENGTH(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

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


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(12)
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(activate(x0)))
   LENGTH1(n__s(x0)) -> LENGTH(s(activate(x0)))
   LENGTH1(n__nil) -> LENGTH(nil)
   LENGTH1(n__cons(x0, x1)) -> LENGTH(cons(activate(x0), x1))
   LENGTH1(x0) -> LENGTH(x0)

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(n__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)
   s(X) -> n__s(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
   activate(X) -> X

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

(13) 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))


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

(14)
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(activate(x0)))
   LENGTH1(n__s(x0)) -> LENGTH(s(activate(x0)))
   LENGTH1(n__cons(x0, x1)) -> LENGTH(cons(activate(x0), x1))
   LENGTH1(x0) -> LENGTH(x0)
   LENGTH1(n__nil) -> LENGTH(n__nil)

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(n__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)
   s(X) -> n__s(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
   activate(X) -> X

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

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

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

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

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(n__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)
   s(X) -> n__s(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
   activate(X) -> X

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

(17) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   LENGTH1(n__s(x0)) -> LENGTH(s(activate(x0)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( LENGTH1_1(x_1) ) = 2x_1 + 2
POL( LENGTH_1(x_1) ) = x_1 + 2
POL( cons_2(x_1, x_2) ) = 2x_2
POL( from_1(x_1) ) = max{0, -2}
POL( s_1(x_1) ) = 1
POL( activate_1(x_1) ) = x_1
POL( n__from_1(x_1) ) = 0
POL( n__s_1(x_1) ) = 1
POL( n__nil ) = 2
POL( nil ) = 2
POL( n__cons_2(x_1, x_2) ) = 2x_2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   activate(n__from(X)) -> from(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
   activate(X) -> X
   from(X) -> cons(X, n__from(n__s(X)))
   from(X) -> n__from(X)
   s(X) -> n__s(X)
   cons(X1, X2) -> n__cons(X1, X2)
   nil -> n__nil


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

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(n__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)
   s(X) -> n__s(X)
   nil -> n__nil
   cons(X1, X2) -> n__cons(X1, X2)
   activate(n__from(X)) -> from(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__nil) -> nil
   activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
   activate(X) -> X

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

(19) 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 = LENGTH(from(X)) evaluates to  t =LENGTH(from(activate(n__s(X))))

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

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Rewriting sequence

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

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

LENGTH(n__cons(X, n__from(n__s(X)))) -> LENGTH1(activate(n__from(n__s(X))))
with rule LENGTH(n__cons(X', Y)) -> LENGTH1(activate(Y)) at position [] and matcher [X' / X, Y / n__from(n__s(X))]

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

LENGTH1(n__from(n__s(X))) -> LENGTH(from(activate(n__s(X))))
with rule LENGTH1(n__from(x0)) -> LENGTH(from(activate(x0)))

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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(20)
NO