YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) RootLabelingProof [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 26 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPOrderProof [EQUIVALENT, 9039 ms]
(10) QDP
(11) PisEmptyProof [EQUIVALENT, 0 ms]
(12) YES


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

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

Q is empty.

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(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

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

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

Q is empty.

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(3) 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.

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

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

Q is empty.

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(5) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1))))))))) -> B_{A_1}(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_{a_1}(a_{b_1}(x1))))))))))))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1))))))))) -> 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_{a_1}(a_{b_1}(x1))))))))))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1))))))))) -> B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1))))))))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1))))))))) -> B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1))))))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1))))))))) -> B_{A_1}(a_{a_1}(a_{b_1}(x1)))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1))))))))) -> B_{A_1}(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_{a_1}(a_{a_1}(x1))))))))))))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1))))))))) -> 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_{a_1}(a_{a_1}(x1))))))))))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1))))))))) -> B_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1))))))))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1))))))))) -> B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1))))))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1))))))))) -> B_{A_1}(a_{a_1}(a_{a_1}(x1)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 8 less nodes.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1))))))))) -> B_{A_1}(a_{a_1}(a_{b_1}(x1)))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1))))))))) -> B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1))))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1))))))))) -> B_{A_1}(a_{a_1}(a_{b_1}(x1)))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1))))))))) -> B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1))))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(B_{A_1}(x_1)) =  	[[-I]] 	 +  	[[0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(a_{a_1}(x_1)) =  	[[-I], [-I], [-I]] 	 +  	[[0A, -I, 0A], [-I, -I, -I], [0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(a_{b_1}(x_1)) =  	[[-I], [-I], [-I]] 	 +  	[[0A, -I, 0A], [-I, -I, 0A], [-I, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(b_{a_1}(x_1)) =  	[[-I], [-I], [-I]] 	 +  	[[0A, 1A, 0A], [0A, 0A, 0A], [-I, 0A, 0A]] 	* 	x_1
>>>


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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


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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(11) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(12)
YES