/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 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) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) NonTerminationLoopProof [COMPLETE, 0 ms]
        (14) NO


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

   f(s(x), x) -> f(s(x), round(x))
   round(0) -> 0
   round(0) -> s(0)
   round(s(0)) -> s(0)
   round(s(s(x))) -> s(s(round(x)))

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(x), x) -> F(s(x), round(x))
   F(s(x), x) -> ROUND(x)
   ROUND(s(s(x))) -> ROUND(x)

The TRS R consists of the following rules:

   f(s(x), x) -> f(s(x), round(x))
   round(0) -> 0
   round(0) -> s(0)
   round(s(0)) -> s(0)
   round(s(s(x))) -> s(s(round(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
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(4)
Complex Obligation (AND)

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

   ROUND(s(s(x))) -> ROUND(x)

The TRS R consists of the following rules:

   f(s(x), x) -> f(s(x), round(x))
   round(0) -> 0
   round(0) -> s(0)
   round(s(0)) -> s(0)
   round(s(s(x))) -> s(s(round(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(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.
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(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ROUND(s(s(x))) -> ROUND(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(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:
*ROUND(s(s(x))) -> ROUND(x)
The graph contains the following edges 1 > 1


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(9)
YES

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

   F(s(x), x) -> F(s(x), round(x))

The TRS R consists of the following rules:

   f(s(x), x) -> f(s(x), round(x))
   round(0) -> 0
   round(0) -> s(0)
   round(s(0)) -> s(0)
   round(s(s(x))) -> s(s(round(x)))

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

(11) 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.
----------------------------------------

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

   F(s(x), x) -> F(s(x), round(x))

The TRS R consists of the following rules:

   round(0) -> 0
   round(0) -> s(0)
   round(s(0)) -> s(0)
   round(s(s(x))) -> s(s(round(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(13) 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 = F(s(0), round(0)) evaluates to  t =F(s(0), round(0))

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

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Rewriting sequence

F(s(0), round(0)) -> F(s(0), 0)
with rule round(0) -> 0 at position [1] and matcher [ ]

F(s(0), 0) -> F(s(0), round(0))
with rule F(s(x), x) -> F(s(x), round(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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(14)
NO