/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 19 ms]
(4) QTRS
(5) AAECC Innermost [EQUIVALENT, 0 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 0 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) UsableRulesProof [EQUIVALENT, 0 ms]
(12) QDP
(13) QReductionProof [EQUIVALENT, 0 ms]
(14) QDP
(15) MNOCProof [EQUIVALENT, 0 ms]
(16) QDP
(17) NonTerminationLoopProof [COMPLETE, 0 ms]
(18) NO


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

   g(X) -> h(activate(X))
   c -> d
   h(n__d) -> g(n__c)
   d -> n__d
   c -> n__c
   activate(n__d) -> d
   activate(n__c) -> c
   activate(X) -> X

Q is empty.

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

(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g(X) -> activate(h(X))
   c'(x) -> d'(x)
   n__d'(h(x)) -> n__c'(g(x))
   d'(x) -> n__d'(x)
   c'(x) -> n__c'(x)
   n__d'(activate(x)) -> d'(x)
   n__c'(activate(x)) -> c'(x)
   activate(X) -> X

Q is empty.

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

(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(activate(x_1)) = 1 + x_1
   POL(c'(x_1)) = 1 + x_1
   POL(d'(x_1)) = 1 + x_1
   POL(g(x_1)) = 1 + x_1
   POL(h(x_1)) = x_1
   POL(n__c'(x_1)) = x_1
   POL(n__d'(x_1)) = 1 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   c'(x) -> n__c'(x)
   n__d'(activate(x)) -> d'(x)
   activate(X) -> X




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

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g(X) -> activate(h(X))
   c'(x) -> d'(x)
   n__d'(h(x)) -> n__c'(g(x))
   d'(x) -> n__d'(x)
   n__c'(activate(x)) -> c'(x)

Q is empty.

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(5) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   g(X) -> activate(h(X))

The TRS R 2 is 
   c'(x) -> d'(x)
   n__d'(h(x)) -> n__c'(g(x))
   d'(x) -> n__d'(x)
   n__c'(activate(x)) -> c'(x)

The signature Sigma is {c'_1, d'_1, n__d'_1, n__c'_1}
----------------------------------------

(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g(X) -> activate(h(X))
   c'(x) -> d'(x)
   n__d'(h(x)) -> n__c'(g(x))
   d'(x) -> n__d'(x)
   n__c'(activate(x)) -> c'(x)

The set Q consists of the following terms:

   g(x0)
   c'(x0)
   n__d'(h(x0))
   d'(x0)
   n__c'(activate(x0))


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

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

   C'(x) -> D'(x)
   N__D'(h(x)) -> N__C'(g(x))
   N__D'(h(x)) -> G(x)
   D'(x) -> N__D'(x)
   N__C'(activate(x)) -> C'(x)

The TRS R consists of the following rules:

   g(X) -> activate(h(X))
   c'(x) -> d'(x)
   n__d'(h(x)) -> n__c'(g(x))
   d'(x) -> n__d'(x)
   n__c'(activate(x)) -> c'(x)

The set Q consists of the following terms:

   g(x0)
   c'(x0)
   n__d'(h(x0))
   d'(x0)
   n__c'(activate(x0))

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:

   D'(x) -> N__D'(x)
   N__D'(h(x)) -> N__C'(g(x))
   N__C'(activate(x)) -> C'(x)
   C'(x) -> D'(x)

The TRS R consists of the following rules:

   g(X) -> activate(h(X))
   c'(x) -> d'(x)
   n__d'(h(x)) -> n__c'(g(x))
   d'(x) -> n__d'(x)
   n__c'(activate(x)) -> c'(x)

The set Q consists of the following terms:

   g(x0)
   c'(x0)
   n__d'(h(x0))
   d'(x0)
   n__c'(activate(x0))

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

(11) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   D'(x) -> N__D'(x)
   N__D'(h(x)) -> N__C'(g(x))
   N__C'(activate(x)) -> C'(x)
   C'(x) -> D'(x)

The TRS R consists of the following rules:

   g(X) -> activate(h(X))

The set Q consists of the following terms:

   g(x0)
   c'(x0)
   n__d'(h(x0))
   d'(x0)
   n__c'(activate(x0))

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

(13) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   c'(x0)
   n__d'(h(x0))
   d'(x0)
   n__c'(activate(x0))


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

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

   D'(x) -> N__D'(x)
   N__D'(h(x)) -> N__C'(g(x))
   N__C'(activate(x)) -> C'(x)
   C'(x) -> D'(x)

The TRS R consists of the following rules:

   g(X) -> activate(h(X))

The set Q consists of the following terms:

   g(x0)

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

(15) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.
----------------------------------------

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

   D'(x) -> N__D'(x)
   N__D'(h(x)) -> N__C'(g(x))
   N__C'(activate(x)) -> C'(x)
   C'(x) -> D'(x)

The TRS R consists of the following rules:

   g(X) -> activate(h(X))

Q is empty.
We have to consider all (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 = N__D'(h(x)) evaluates to  t =N__D'(h(x))

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

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Rewriting sequence

N__D'(h(x)) -> N__C'(g(x))
with rule N__D'(h(x')) -> N__C'(g(x')) at position [] and matcher [x' / x]

N__C'(g(x)) -> N__C'(activate(h(x)))
with rule g(X) -> activate(h(X)) at position [0] and matcher [X / x]

N__C'(activate(h(x))) -> C'(h(x))
with rule N__C'(activate(x')) -> C'(x') at position [] and matcher [x' / h(x)]

C'(h(x)) -> D'(h(x))
with rule C'(x') -> D'(x') at position [] and matcher [x' / h(x)]

D'(h(x)) -> N__D'(h(x))
with rule D'(x) -> N__D'(x)

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