/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 68 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) TransformationProof [EQUIVALENT, 0 ms]
(10) QDP
(11) TransformationProof [EQUIVALENT, 0 ms]
(12) QDP
(13) TransformationProof [EQUIVALENT, 0 ms]
(14) QDP
(15) DependencyGraphProof [EQUIVALENT, 0 ms]
(16) QDP
(17) NonTerminationLoopProof [COMPLETE, 11 ms]
(18) NO


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

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   head(cons(X, XS)) -> X
   tail(cons(X, XS)) -> activate(XS)
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   activate(X) -> X

Q is empty.

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

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

   POL(0) = 0
   POL(activate(x_1)) = x_1
   POL(cons(x_1, x_2)) = 2*x_1 + x_2
   POL(head(x_1)) = 2*x_1
   POL(incr(x_1)) = 2*x_1
   POL(n__incr(x_1)) = 2*x_1
   POL(n__nats) = 0
   POL(n__odds) = 0
   POL(nats) = 0
   POL(odds) = 0
   POL(pairs) = 0
   POL(s(x_1)) = 2*x_1
   POL(tail(x_1)) = 2 + 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   tail(cons(X, XS)) -> activate(XS)




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

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

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   head(cons(X, XS)) -> X
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   activate(X) -> X

Q is empty.

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

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

   POL(0) = 0
   POL(activate(x_1)) = x_1
   POL(cons(x_1, x_2)) = 2*x_1 + x_2
   POL(head(x_1)) = 1 + 2*x_1
   POL(incr(x_1)) = 2*x_1
   POL(n__incr(x_1)) = 2*x_1
   POL(n__nats) = 0
   POL(n__odds) = 0
   POL(nats) = 0
   POL(odds) = 0
   POL(pairs) = 0
   POL(s(x_1)) = 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   head(cons(X, XS)) -> X




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

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

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   activate(X) -> X

Q is empty.

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

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

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

   ODDS -> INCR(pairs)
   ODDS -> PAIRS
   INCR(cons(X, XS)) -> ACTIVATE(XS)
   ACTIVATE(n__incr(X)) -> INCR(activate(X))
   ACTIVATE(n__incr(X)) -> ACTIVATE(X)
   ACTIVATE(n__nats) -> NATS
   ACTIVATE(n__odds) -> ODDS

The TRS R consists of the following rules:

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   activate(X) -> X

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

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

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

   INCR(cons(X, XS)) -> ACTIVATE(XS)
   ACTIVATE(n__incr(X)) -> INCR(activate(X))
   ACTIVATE(n__incr(X)) -> ACTIVATE(X)
   ACTIVATE(n__odds) -> ODDS
   ODDS -> INCR(pairs)

The TRS R consists of the following rules:

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   activate(X) -> X

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

(9) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule ODDS -> INCR(pairs) at position [0] we obtained the following new rules [LPAR04]:

   (ODDS -> INCR(cons(0, n__incr(n__odds))),ODDS -> INCR(cons(0, n__incr(n__odds))))


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

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

   INCR(cons(X, XS)) -> ACTIVATE(XS)
   ACTIVATE(n__incr(X)) -> INCR(activate(X))
   ACTIVATE(n__incr(X)) -> ACTIVATE(X)
   ACTIVATE(n__odds) -> ODDS
   ODDS -> INCR(cons(0, n__incr(n__odds)))

The TRS R consists of the following rules:

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   activate(X) -> X

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

(11) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule ACTIVATE(n__incr(X)) -> INCR(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

   (ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0))),ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0))))
   (ACTIVATE(n__incr(n__nats)) -> INCR(nats),ACTIVATE(n__incr(n__nats)) -> INCR(nats))
   (ACTIVATE(n__incr(n__odds)) -> INCR(odds),ACTIVATE(n__incr(n__odds)) -> INCR(odds))
   (ACTIVATE(n__incr(x0)) -> INCR(x0),ACTIVATE(n__incr(x0)) -> INCR(x0))


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

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

   INCR(cons(X, XS)) -> ACTIVATE(XS)
   ACTIVATE(n__incr(X)) -> ACTIVATE(X)
   ACTIVATE(n__odds) -> ODDS
   ODDS -> INCR(cons(0, n__incr(n__odds)))
   ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0)))
   ACTIVATE(n__incr(n__nats)) -> INCR(nats)
   ACTIVATE(n__incr(n__odds)) -> INCR(odds)
   ACTIVATE(n__incr(x0)) -> INCR(x0)

The TRS R consists of the following rules:

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   activate(X) -> X

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

(13) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule ACTIVATE(n__incr(n__nats)) -> INCR(nats) at position [0] we obtained the following new rules [LPAR04]:

   (ACTIVATE(n__incr(n__nats)) -> INCR(cons(0, n__incr(n__nats))),ACTIVATE(n__incr(n__nats)) -> INCR(cons(0, n__incr(n__nats))))
   (ACTIVATE(n__incr(n__nats)) -> INCR(n__nats),ACTIVATE(n__incr(n__nats)) -> INCR(n__nats))


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

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

   INCR(cons(X, XS)) -> ACTIVATE(XS)
   ACTIVATE(n__incr(X)) -> ACTIVATE(X)
   ACTIVATE(n__odds) -> ODDS
   ODDS -> INCR(cons(0, n__incr(n__odds)))
   ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0)))
   ACTIVATE(n__incr(n__odds)) -> INCR(odds)
   ACTIVATE(n__incr(x0)) -> INCR(x0)
   ACTIVATE(n__incr(n__nats)) -> INCR(cons(0, n__incr(n__nats)))
   ACTIVATE(n__incr(n__nats)) -> INCR(n__nats)

The TRS R consists of the following rules:

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   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:

   ACTIVATE(n__incr(X)) -> ACTIVATE(X)
   ACTIVATE(n__odds) -> ODDS
   ODDS -> INCR(cons(0, n__incr(n__odds)))
   INCR(cons(X, XS)) -> ACTIVATE(XS)
   ACTIVATE(n__incr(n__incr(x0))) -> INCR(incr(activate(x0)))
   ACTIVATE(n__incr(n__odds)) -> INCR(odds)
   ACTIVATE(n__incr(x0)) -> INCR(x0)
   ACTIVATE(n__incr(n__nats)) -> INCR(cons(0, n__incr(n__nats)))

The TRS R consists of the following rules:

   nats -> cons(0, n__incr(n__nats))
   pairs -> cons(0, n__incr(n__odds))
   odds -> incr(pairs)
   incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS)))
   incr(X) -> n__incr(X)
   nats -> n__nats
   odds -> n__odds
   activate(n__incr(X)) -> incr(activate(X))
   activate(n__nats) -> nats
   activate(n__odds) -> odds
   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 = ACTIVATE(n__incr(n__nats)) evaluates to  t =ACTIVATE(n__incr(n__nats))

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

--------------------------------------------------------------------------------
Rewriting sequence

ACTIVATE(n__incr(n__nats)) -> INCR(cons(0, n__incr(n__nats)))
with rule ACTIVATE(n__incr(n__nats)) -> INCR(cons(0, n__incr(n__nats))) at position [] and matcher [ ]

INCR(cons(0, n__incr(n__nats))) -> ACTIVATE(n__incr(n__nats))
with rule INCR(cons(X, XS)) -> ACTIVATE(XS)

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