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


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 29 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 1 ms]
(4) QDP
(5) SemLabProof [SOUND, 169 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) MRRProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) PisEmptyProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesReductionPairsProof [EQUIVALENT, 2 ms]
        (18) QDP
        (19) MRRProof [EQUIVALENT, 6 ms]
        (20) QDP
        (21) PisEmptyProof [EQUIVALENT, 0 ms]
        (22) YES


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

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

   a(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> a(a(b(a(b(a(a(b(a(b(a(a(b(a(x1))))))))))))))

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:

   A(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> A(a(b(a(b(a(a(b(a(b(a(a(b(a(x1))))))))))))))
   A(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> A(b(a(b(a(a(b(a(b(a(a(b(a(x1)))))))))))))
   A(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> A(b(a(a(b(a(b(a(a(b(a(x1)))))))))))
   A(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> A(a(b(a(b(a(a(b(a(x1)))))))))
   A(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> A(b(a(b(a(a(b(a(x1))))))))
   A(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> A(b(a(a(b(a(x1))))))
   A(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> A(a(b(a(x1))))
   A(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> A(b(a(x1)))
   A(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> A(x1)

The TRS R consists of the following rules:

   a(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> a(a(b(a(b(a(a(b(a(b(a(a(b(a(x1))))))))))))))

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 7 less nodes.
----------------------------------------

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

   A(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> A(x1)
   A(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> A(a(b(a(x1))))

The TRS R consists of the following rules:

   a(a(b(a(a(b(a(b(a(a(b(x1))))))))))) -> a(a(b(a(b(a(a(b(a(b(a(a(b(a(x1))))))))))))))

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

(5) SemLabProof (SOUND)
We found the following model for the rules of the TRSs R and P.
Interpretation over the domain with elements from 0 to 1.
a: 1 + x<sub>0</sub>
b: x<sub>0</sub>
A: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

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

   A.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> A.0(x1)
   A.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> A.0(a.1(b.1(a.0(x1))))
   A.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> A.1(a.0(b.0(a.1(x1))))
   A.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> A.1(x1)

The TRS R consists of the following rules:

   a.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(x1))))))))))))))
   a.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(x1))))))))))))))

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 2 SCCs.
----------------------------------------

(8)
Complex Obligation (AND)

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

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> A.1(x1)
   A.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> A.1(a.0(b.0(a.1(x1))))

The TRS R consists of the following rules:

   a.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(x1))))))))))))))
   a.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(x1))))))))))))))

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

(10) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

   a.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(x1))))))))))))))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(A.1(x_1)) = x_1
   POL(a.0(x_1)) = x_1
   POL(a.1(x_1)) = x_1
   POL(b.0(x_1)) = x_1
   POL(b.1(x_1)) = x_1


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> A.1(x1)
   A.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> A.1(a.0(b.0(a.1(x1))))

The TRS R consists of the following rules:

   a.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(x1))))))))))))))

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

(12) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   A.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> A.1(x1)
   A.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> A.1(a.0(b.0(a.1(x1))))


Used ordering: Polynomial interpretation [POLO]:

   POL(A.1(x_1)) = x_1
   POL(a.0(x_1)) = x_1
   POL(a.1(x_1)) = x_1
   POL(b.0(x_1)) = x_1
   POL(b.1(x_1)) = 1 + x_1


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

(13)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   a.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(x1))))))))))))))

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

(14) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(15)
YES

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

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

   A.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> A.0(a.1(b.1(a.0(x1))))
   A.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> A.0(x1)

The TRS R consists of the following rules:

   a.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(x1))))))))))))))
   a.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(x1))))))))))))))

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

(17) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

   a.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(x1))))))))))) -> a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(x1))))))))))))))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(A.0(x_1)) = x_1
   POL(a.0(x_1)) = x_1
   POL(a.1(x_1)) = x_1
   POL(b.0(x_1)) = x_1
   POL(b.1(x_1)) = x_1


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

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

   A.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> A.0(a.1(b.1(a.0(x1))))
   A.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> A.0(x1)

The TRS R consists of the following rules:

   a.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(x1))))))))))))))

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

(19) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   A.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> A.0(a.1(b.1(a.0(x1))))
   A.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> A.0(x1)


Used ordering: Polynomial interpretation [POLO]:

   POL(A.0(x_1)) = x_1
   POL(a.0(x_1)) = x_1
   POL(a.1(x_1)) = x_1
   POL(b.0(x_1)) = 1 + x_1
   POL(b.1(x_1)) = x_1


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

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

   a.0(a.1(b.1(a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(x1))))))))))) -> a.0(a.1(b.1(a.0(b.0(a.1(a.0(b.0(a.1(b.1(a.0(a.1(b.1(a.0(x1))))))))))))))

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

(21) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(22)
YES