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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 2 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(10) YES


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

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

Q is empty.

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(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

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

   A_{B_1}(b_{a_1}(a_{a_1}(x1))) -> A_{B_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))))
   A_{B_1}(b_{a_1}(a_{a_1}(x1))) -> B_{A_1}(a_{a_1}(a_{a_1}(x1)))
   A_{B_1}(b_{a_1}(a_{b_1}(x1))) -> A_{B_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))))
   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}(x1)))) -> B_{A_1}(a_{b_1}(b_{a_1}(x1)))
   B_{A_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> B_{A_1}(a_{b_1}(b_{b_1}(x1)))

The TRS R consists of the following rules:

   a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1))))))
   a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1))))))
   b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{a_1}(x1)))
   b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_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 5 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}(x1)))) -> 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_{a_1}(x1))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1))))))
   a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1))))))
   b_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{a_1}(x1)))
   b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(9) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*B_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> B_{A_1}(a_{b_1}(b_{a_1}(x1)))
The graph contains the following edges 1 > 1


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(10)
YES