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


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


NO
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QDP
(5) UsableRulesProof [EQUIVALENT, 0 ms]
(6) QDP
(7) TransformationProof [EQUIVALENT, 0 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) UsableRulesProof [EQUIVALENT, 0 ms]
(12) QDP
(13) TransformationProof [EQUIVALENT, 0 ms]
(14) QDP
(15) DependencyGraphProof [EQUIVALENT, 0 ms]
(16) QDP
(17) TransformationProof [EQUIVALENT, 0 ms]
(18) QDP
(19) TransformationProof [EQUIVALENT, 0 ms]
(20) QDP
(21) TransformationProof [EQUIVALENT, 0 ms]
(22) QDP
(23) DependencyGraphProof [EQUIVALENT, 0 ms]
(24) QDP
(25) TransformationProof [EQUIVALENT, 0 ms]
(26) QDP
(27) UsableRulesProof [EQUIVALENT, 0 ms]
(28) QDP
(29) MNOCProof [EQUIVALENT, 0 ms]
(30) QDP
(31) TransformationProof [EQUIVALENT, 0 ms]
(32) QDP
(33) DependencyGraphProof [EQUIVALENT, 0 ms]
(34) QDP
(35) UsableRulesProof [EQUIVALENT, 0 ms]
(36) QDP
(37) QReductionProof [EQUIVALENT, 0 ms]
(38) QDP
(39) TransformationProof [EQUIVALENT, 0 ms]
(40) QDP
(41) DependencyGraphProof [EQUIVALENT, 0 ms]
(42) QDP
(43) NonTerminationLoopProof [COMPLETE, 0 ms]
(44) NO


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

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

   f(n__a, X, X) -> f(activate(X), b, n__b)
   b -> a
   a -> n__a
   b -> n__b
   activate(n__a) -> a
   activate(n__b) -> b
   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:

   F(n__a, X, X) -> F(activate(X), b, n__b)
   F(n__a, X, X) -> ACTIVATE(X)
   F(n__a, X, X) -> B
   B -> A
   ACTIVATE(n__a) -> A
   ACTIVATE(n__b) -> B

The TRS R consists of the following rules:

   f(n__a, X, X) -> f(activate(X), b, n__b)
   b -> a
   a -> n__a
   b -> n__b
   activate(n__a) -> a
   activate(n__b) -> b
   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 5 less nodes.
----------------------------------------

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

   F(n__a, X, X) -> F(activate(X), b, n__b)

The TRS R consists of the following rules:

   f(n__a, X, X) -> f(activate(X), b, n__b)
   b -> a
   a -> n__a
   b -> n__b
   activate(n__a) -> a
   activate(n__b) -> b
   activate(X) -> X

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

(5) 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.
----------------------------------------

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

   F(n__a, X, X) -> F(activate(X), b, n__b)

The TRS R consists of the following rules:

   activate(n__a) -> a
   activate(n__b) -> b
   activate(X) -> X
   b -> a
   b -> n__b
   a -> n__a

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

(7) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule F(n__a, X, X) -> F(activate(X), b, n__b) at position [0] we obtained the following new rules [LPAR04]:

   (F(n__a, n__a, n__a) -> F(a, b, n__b),F(n__a, n__a, n__a) -> F(a, b, n__b))
   (F(n__a, n__b, n__b) -> F(b, b, n__b),F(n__a, n__b, n__b) -> F(b, b, n__b))
   (F(n__a, x0, x0) -> F(x0, b, n__b),F(n__a, x0, x0) -> F(x0, b, n__b))


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

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

   F(n__a, n__a, n__a) -> F(a, b, n__b)
   F(n__a, n__b, n__b) -> F(b, b, n__b)
   F(n__a, x0, x0) -> F(x0, b, n__b)

The TRS R consists of the following rules:

   activate(n__a) -> a
   activate(n__b) -> b
   activate(X) -> X
   b -> a
   b -> n__b
   a -> n__a

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:

   F(n__a, n__b, n__b) -> F(b, b, n__b)
   F(n__a, x0, x0) -> F(x0, b, n__b)

The TRS R consists of the following rules:

   activate(n__a) -> a
   activate(n__b) -> b
   activate(X) -> X
   b -> a
   b -> n__b
   a -> n__a

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(n__a, n__b, n__b) -> F(b, b, n__b)
   F(n__a, x0, x0) -> F(x0, b, n__b)

The TRS R consists of the following rules:

   b -> a
   b -> n__b
   a -> n__a

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

(13) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule F(n__a, n__b, n__b) -> F(b, b, n__b) at position [0] we obtained the following new rules [LPAR04]:

   (F(n__a, n__b, n__b) -> F(a, b, n__b),F(n__a, n__b, n__b) -> F(a, b, n__b))
   (F(n__a, n__b, n__b) -> F(n__b, b, n__b),F(n__a, n__b, n__b) -> F(n__b, b, n__b))


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

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

   F(n__a, x0, x0) -> F(x0, b, n__b)
   F(n__a, n__b, n__b) -> F(a, b, n__b)
   F(n__a, n__b, n__b) -> F(n__b, b, n__b)

The TRS R consists of the following rules:

   b -> a
   b -> n__b
   a -> n__a

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:

   F(n__a, x0, x0) -> F(x0, b, n__b)
   F(n__a, n__b, n__b) -> F(a, b, n__b)

The TRS R consists of the following rules:

   b -> a
   b -> n__b
   a -> n__a

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

(17) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule F(n__a, x0, x0) -> F(x0, b, n__b) at position [] we obtained the following new rules [LPAR04]:

   (F(n__a, y0, y0) -> F(y0, a, n__b),F(n__a, y0, y0) -> F(y0, a, n__b))
   (F(n__a, y0, y0) -> F(y0, n__b, n__b),F(n__a, y0, y0) -> F(y0, n__b, n__b))


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

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

   F(n__a, n__b, n__b) -> F(a, b, n__b)
   F(n__a, y0, y0) -> F(y0, a, n__b)
   F(n__a, y0, y0) -> F(y0, n__b, n__b)

The TRS R consists of the following rules:

   b -> a
   b -> n__b
   a -> n__a

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

(19) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule F(n__a, n__b, n__b) -> F(a, b, n__b) at position [0] we obtained the following new rules [LPAR04]:

   (F(n__a, n__b, n__b) -> F(n__a, b, n__b),F(n__a, n__b, n__b) -> F(n__a, b, n__b))


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

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, y0, y0) -> F(y0, a, n__b)
   F(n__a, y0, y0) -> F(y0, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__a, b, n__b)

The TRS R consists of the following rules:

   b -> a
   b -> n__b
   a -> n__a

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

(21) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule F(n__a, y0, y0) -> F(y0, a, n__b) at position [] we obtained the following new rules [LPAR04]:

   (F(n__a, y0, y0) -> F(y0, n__a, n__b),F(n__a, y0, y0) -> F(y0, n__a, n__b))


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

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, y0, y0) -> F(y0, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__a, b, n__b)
   F(n__a, y0, y0) -> F(y0, n__a, n__b)

The TRS R consists of the following rules:

   b -> a
   b -> n__b
   a -> n__a

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

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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, y0, y0) -> F(y0, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__a, b, n__b)

The TRS R consists of the following rules:

   b -> a
   b -> n__b
   a -> n__a

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

(25) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule F(n__a, n__b, n__b) -> F(n__a, b, n__b) at position [] we obtained the following new rules [LPAR04]:

   (F(n__a, n__b, n__b) -> F(n__a, a, n__b),F(n__a, n__b, n__b) -> F(n__a, a, n__b))
   (F(n__a, n__b, n__b) -> F(n__a, n__b, n__b),F(n__a, n__b, n__b) -> F(n__a, n__b, n__b))


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

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, y0, y0) -> F(y0, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__a, a, n__b)
   F(n__a, n__b, n__b) -> F(n__a, n__b, n__b)

The TRS R consists of the following rules:

   b -> a
   b -> n__b
   a -> n__a

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

(27) 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.
----------------------------------------

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, y0, y0) -> F(y0, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__a, a, n__b)
   F(n__a, n__b, n__b) -> F(n__a, n__b, n__b)

The TRS R consists of the following rules:

   a -> n__a

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

(29) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, y0, y0) -> F(y0, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__a, a, n__b)
   F(n__a, n__b, n__b) -> F(n__a, n__b, n__b)

The TRS R consists of the following rules:

   a -> n__a

The set Q consists of the following terms:

   a

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

(31) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule F(n__a, n__b, n__b) -> F(n__a, a, n__b) at position [1] we obtained the following new rules [LPAR04]:

   (F(n__a, n__b, n__b) -> F(n__a, n__a, n__b),F(n__a, n__b, n__b) -> F(n__a, n__a, n__b))


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

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, y0, y0) -> F(y0, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__a, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__a, n__a, n__b)

The TRS R consists of the following rules:

   a -> n__a

The set Q consists of the following terms:

   a

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

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

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, y0, y0) -> F(y0, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__a, n__b, n__b)

The TRS R consists of the following rules:

   a -> n__a

The set Q consists of the following terms:

   a

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

(35) 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.
----------------------------------------

(36)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, y0, y0) -> F(y0, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__a, n__b, n__b)

R is empty.
The set Q consists of the following terms:

   a

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

(37) 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].

   a


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

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, y0, y0) -> F(y0, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__a, n__b, n__b)

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

(39) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule F(n__a, y0, y0) -> F(y0, n__b, n__b) we obtained the following new rules [LPAR04]:

   (F(n__a, n__b, n__b) -> F(n__b, n__b, n__b),F(n__a, n__b, n__b) -> F(n__b, n__b, n__b))


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

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, n__b, n__b) -> F(n__a, n__b, n__b)
   F(n__a, n__b, n__b) -> F(n__b, n__b, n__b)

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

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

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(n__a, n__b, n__b) -> F(n__a, n__b, n__b)

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

(43) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = F(n__a, n__b, n__b) evaluates to  t =F(n__a, n__b, n__b)

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

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

The DP semiunifies directly so there is only one rewrite step from F(n__a, n__b, n__b) to F(n__a, n__b, n__b).




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

(44)
NO