/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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) RootLabelingProof [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 8 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) SemLabProof [SOUND, 3313 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) AND
    (11) QDP
        (12) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) PisEmptyProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) PisEmptyProof [EQUIVALENT, 0 ms]
        (20) YES


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

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

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

Q is empty.

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

(1) RootLabelingProof (EQUIVALENT)
We used plain root labeling [ROOTLAB] with the following heuristic:
LabelAll: All function symbols get labeled

As Q is empty the root labeling was sound AND complete.

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

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

   a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))))))))))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))))))))))
   a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))))))))))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))))))))))

Q is empty.

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

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

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

   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))))))))))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))))))))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))))))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(x1))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))))))))))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))))))))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))))))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))))))))))) -> A_{B_1}(b_{b_1}(x1))

The TRS R consists of the following rules:

   a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))))))))))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))))))))))
   a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))))))))))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))))))))))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 10 less nodes.
----------------------------------------

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

   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(x1))
   A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))))))))))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))

The TRS R consists of the following rules:

   a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))))))))))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))))))))))))
   a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))))))))))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))))))))))))

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

(7) 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.
b_{b_1}: 1 + x<sub>0</sub>
b_{a_1}: x<sub>0</sub>
A_{B_1}: 0
a_{b_1}: x<sub>0</sub>
a_{a_1}: 1 + x<sub>0</sub>
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
----------------------------------------

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

   A_{B_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(x1)))))))))))) -> A_{B_1}.0(b_{a_1}.0(x1))
   A_{B_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(x1)))))))))))) -> A_{B_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(x1))))
   A_{B_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(x1)))))))))))) -> A_{B_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(x1))))
   A_{B_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(x1)))))))))))) -> A_{B_1}.1(b_{a_1}.1(x1))

The TRS R consists of the following rules:

   a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(x1)))))))))))) -> a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(x1))))))))))))))
   a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(x1)))))))))))) -> a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(x1))))))))))))))
   a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(x1)))))))))))) -> a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{b_1}.0(x1))))))))))))))
   a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(x1)))))))))))) -> a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{b_1}.1(x1))))))))))))))

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

(10)
Complex Obligation (AND)

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

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

   A_{B_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(x1)))))))))))) -> A_{B_1}.1(b_{a_1}.1(x1))
   A_{B_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(x1)))))))))))) -> A_{B_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(x1))))

The TRS R consists of the following rules:

   a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(x1)))))))))))) -> a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(x1))))))))))))))
   a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(x1)))))))))))) -> a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(x1))))))))))))))
   a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(x1)))))))))))) -> a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{b_1}.0(x1))))))))))))))
   a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(x1)))))))))))) -> a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{b_1}.1(x1))))))))))))))

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

(12) 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.

The following dependency pairs can be deleted:

   A_{B_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(x1)))))))))))) -> A_{B_1}.1(b_{a_1}.1(x1))
   A_{B_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(x1)))))))))))) -> A_{B_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(x1))))
The following rules are removed from R:

   a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(x1)))))))))))) -> a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(x1))))))))))))))
   a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(x1)))))))))))) -> a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{b_1}.1(x1))))))))))))))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(A_{B_1}.1(x_1)) = x_1
   POL(a_{a_1}.0(x_1)) = x_1
   POL(a_{a_1}.1(x_1)) = 1 + x_1
   POL(a_{b_1}.0(x_1)) = 1 + x_1
   POL(a_{b_1}.1(x_1)) = x_1
   POL(b_{a_1}.0(x_1)) = x_1
   POL(b_{a_1}.1(x_1)) = x_1
   POL(b_{b_1}.0(x_1)) = x_1


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

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

   a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(x1)))))))))))) -> a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(x1))))))))))))))
   a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(x1)))))))))))) -> a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{b_1}.0(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

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

   A_{B_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(x1)))))))))))) -> A_{B_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(x1))))
   A_{B_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(x1)))))))))))) -> A_{B_1}.0(b_{a_1}.0(x1))

The TRS R consists of the following rules:

   a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(x1)))))))))))) -> a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(x1))))))))))))))
   a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(x1)))))))))))) -> a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(x1))))))))))))))
   a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(x1)))))))))))) -> a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{b_1}.0(x1))))))))))))))
   a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(x1)))))))))))) -> a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{b_1}.1(x1))))))))))))))

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

The following dependency pairs can be deleted:

   A_{B_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(x1)))))))))))) -> A_{B_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(x1))))
   A_{B_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(x1)))))))))))) -> A_{B_1}.0(b_{a_1}.0(x1))
The following rules are removed from R:

   a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(x1)))))))))))) -> a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(x1))))))))))))))
   a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(x1)))))))))))) -> a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{b_1}.0(x1))))))))))))))
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(A_{B_1}.0(x_1)) = x_1
   POL(a_{a_1}.0(x_1)) = 1 + x_1
   POL(a_{a_1}.1(x_1)) = x_1
   POL(a_{b_1}.0(x_1)) = x_1
   POL(a_{b_1}.1(x_1)) = 1 + x_1
   POL(b_{a_1}.0(x_1)) = x_1
   POL(b_{a_1}.1(x_1)) = x_1
   POL(b_{b_1}.1(x_1)) = x_1


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

   a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(x1)))))))))))) -> a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(x1))))))))))))))
   a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(x1)))))))))))) -> a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{b_1}.1(b_{a_1}.1(a_{a_1}.0(a_{b_1}.0(b_{a_1}.0(a_{b_1}.0(b_{b_1}.1(x1))))))))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(19) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(20)
YES