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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 151 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) TRUE


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

   0(1(0(2(x1)))) -> 2(2(2(x1)))
   0(1(1(1(x1)))) -> 3(1(0(x1)))
   4(5(1(4(x1)))) -> 1(5(4(4(x1))))
   1(2(0(4(1(x1))))) -> 2(5(2(2(x1))))
   1(5(1(0(3(x1))))) -> 5(5(0(3(x1))))
   1(3(5(1(0(4(x1)))))) -> 4(0(0(0(1(4(x1))))))
   3(3(1(4(5(4(5(5(x1)))))))) -> 3(5(3(2(0(5(2(x1)))))))
   4(2(5(3(1(0(0(5(2(x1))))))))) -> 4(4(1(5(3(5(3(2(x1))))))))
   2(2(1(1(0(2(0(1(2(3(x1)))))))))) -> 5(4(4(5(2(5(5(2(3(x1)))))))))
   2(2(2(3(3(4(3(3(5(4(5(x1))))))))))) -> 5(5(4(2(5(3(5(3(2(0(5(x1)))))))))))
   4(2(1(1(5(1(0(0(0(2(5(x1))))))))))) -> 5(5(0(3(0(1(4(5(2(5(x1))))))))))
   5(4(5(4(2(0(2(2(3(0(0(2(x1)))))))))))) -> 5(3(4(5(0(3(2(1(1(2(2(x1)))))))))))
   0(2(1(1(3(1(5(0(5(2(3(1(4(x1))))))))))))) -> 3(0(1(5(1(4(4(5(0(5(1(1(4(x1)))))))))))))
   1(1(2(4(2(2(2(1(1(3(4(1(2(x1))))))))))))) -> 1(5(0(3(4(1(0(5(2(2(2(3(x1))))))))))))
   1(4(0(0(2(4(0(3(3(2(0(3(1(x1))))))))))))) -> 1(5(0(5(0(2(0(5(1(0(1(2(5(5(x1))))))))))))))
   4(1(4(1(3(0(0(2(1(5(4(1(0(0(x1)))))))))))))) -> 2(3(0(2(5(4(4(3(0(4(1(2(0(x1)))))))))))))
   4(2(4(4(0(0(0(4(0(3(5(0(3(3(x1)))))))))))))) -> 5(1(0(1(0(1(3(5(3(5(5(0(0(3(x1))))))))))))))
   1(2(5(3(3(0(2(2(5(3(2(3(3(3(2(1(x1)))))))))))))))) -> 1(1(0(0(4(1(2(1(0(5(0(0(3(5(0(4(1(x1)))))))))))))))))
   3(1(2(5(0(3(4(3(1(5(4(1(5(2(0(5(x1)))))))))))))))) -> 3(1(0(4(2(0(5(2(4(4(2(2(1(1(1(5(x1))))))))))))))))
   1(3(4(5(1(0(0(3(1(2(4(2(3(5(2(0(4(1(x1)))))))))))))))))) -> 1(3(0(0(4(2(2(5(3(1(0(1(2(1(5(0(1(4(1(x1)))))))))))))))))))
   1(0(3(3(2(5(0(0(3(0(3(2(4(1(4(0(2(4(2(x1))))))))))))))))))) -> 4(3(4(2(1(4(2(4(3(3(2(2(2(1(1(1(5(3(2(x1)))))))))))))))))))
   5(3(2(1(0(1(3(1(3(3(0(0(3(2(5(3(0(3(0(x1))))))))))))))))))) -> 5(1(0(3(2(4(0(0(3(2(1(5(1(3(0(5(1(3(1(x1)))))))))))))))))))
   0(2(2(3(3(5(0(0(5(0(3(1(3(0(1(2(1(5(5(1(x1)))))))))))))))))))) -> 0(0(3(2(5(2(5(2(3(2(1(2(5(4(3(4(5(0(4(x1)))))))))))))))))))
   1(0(4(2(3(3(5(4(3(5(0(2(0(4(5(0(2(0(2(4(x1)))))))))))))))))))) -> 4(5(4(5(4(1(1(2(5(0(4(3(1(5(4(3(1(5(4(0(4(x1)))))))))))))))))))))
   1(4(2(3(1(3(4(2(4(1(5(1(4(0(4(5(2(0(0(3(4(x1))))))))))))))))))))) -> 4(3(4(1(3(5(1(1(4(3(1(5(1(3(1(2(4(2(1(3(x1))))))))))))))))))))

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0(x_1)) = 235 + x_1
   POL(1(x_1)) = 191 + x_1
   POL(2(x_1)) = 157 + x_1
   POL(3(x_1)) = 381 + x_1
   POL(4(x_1)) = 237 + x_1
   POL(5(x_1)) = 136 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   0(1(0(2(x1)))) -> 2(2(2(x1)))
   0(1(1(1(x1)))) -> 3(1(0(x1)))
   1(2(0(4(1(x1))))) -> 2(5(2(2(x1))))
   1(5(1(0(3(x1))))) -> 5(5(0(3(x1))))
   1(3(5(1(0(4(x1)))))) -> 4(0(0(0(1(4(x1))))))
   3(3(1(4(5(4(5(5(x1)))))))) -> 3(5(3(2(0(5(2(x1)))))))
   4(2(5(3(1(0(0(5(2(x1))))))))) -> 4(4(1(5(3(5(3(2(x1))))))))
   2(2(1(1(0(2(0(1(2(3(x1)))))))))) -> 5(4(4(5(2(5(5(2(3(x1)))))))))
   2(2(2(3(3(4(3(3(5(4(5(x1))))))))))) -> 5(5(4(2(5(3(5(3(2(0(5(x1)))))))))))
   4(2(1(1(5(1(0(0(0(2(5(x1))))))))))) -> 5(5(0(3(0(1(4(5(2(5(x1))))))))))
   5(4(5(4(2(0(2(2(3(0(0(2(x1)))))))))))) -> 5(3(4(5(0(3(2(1(1(2(2(x1)))))))))))
   0(2(1(1(3(1(5(0(5(2(3(1(4(x1))))))))))))) -> 3(0(1(5(1(4(4(5(0(5(1(1(4(x1)))))))))))))
   1(1(2(4(2(2(2(1(1(3(4(1(2(x1))))))))))))) -> 1(5(0(3(4(1(0(5(2(2(2(3(x1))))))))))))
   1(4(0(0(2(4(0(3(3(2(0(3(1(x1))))))))))))) -> 1(5(0(5(0(2(0(5(1(0(1(2(5(5(x1))))))))))))))
   4(1(4(1(3(0(0(2(1(5(4(1(0(0(x1)))))))))))))) -> 2(3(0(2(5(4(4(3(0(4(1(2(0(x1)))))))))))))
   4(2(4(4(0(0(0(4(0(3(5(0(3(3(x1)))))))))))))) -> 5(1(0(1(0(1(3(5(3(5(5(0(0(3(x1))))))))))))))
   1(2(5(3(3(0(2(2(5(3(2(3(3(3(2(1(x1)))))))))))))))) -> 1(1(0(0(4(1(2(1(0(5(0(0(3(5(0(4(1(x1)))))))))))))))))
   3(1(2(5(0(3(4(3(1(5(4(1(5(2(0(5(x1)))))))))))))))) -> 3(1(0(4(2(0(5(2(4(4(2(2(1(1(1(5(x1))))))))))))))))
   1(3(4(5(1(0(0(3(1(2(4(2(3(5(2(0(4(1(x1)))))))))))))))))) -> 1(3(0(0(4(2(2(5(3(1(0(1(2(1(5(0(1(4(1(x1)))))))))))))))))))
   1(0(3(3(2(5(0(0(3(0(3(2(4(1(4(0(2(4(2(x1))))))))))))))))))) -> 4(3(4(2(1(4(2(4(3(3(2(2(2(1(1(1(5(3(2(x1)))))))))))))))))))
   5(3(2(1(0(1(3(1(3(3(0(0(3(2(5(3(0(3(0(x1))))))))))))))))))) -> 5(1(0(3(2(4(0(0(3(2(1(5(1(3(0(5(1(3(1(x1)))))))))))))))))))
   0(2(2(3(3(5(0(0(5(0(3(1(3(0(1(2(1(5(5(1(x1)))))))))))))))))))) -> 0(0(3(2(5(2(5(2(3(2(1(2(5(4(3(4(5(0(4(x1)))))))))))))))))))
   1(0(4(2(3(3(5(4(3(5(0(2(0(4(5(0(2(0(2(4(x1)))))))))))))))))))) -> 4(5(4(5(4(1(1(2(5(0(4(3(1(5(4(3(1(5(4(0(4(x1)))))))))))))))))))))
   1(4(2(3(1(3(4(2(4(1(5(1(4(0(4(5(2(0(0(3(4(x1))))))))))))))))))))) -> 4(3(4(1(3(5(1(1(4(3(1(5(1(3(1(2(4(2(1(3(x1))))))))))))))))))))




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

   4(5(1(4(x1)))) -> 1(5(4(4(x1))))

Q is empty.

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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   4^1(5(1(4(x1)))) -> 4^1(4(x1))

The TRS R consists of the following rules:

   4(5(1(4(x1)))) -> 1(5(4(4(x1))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(6)
TRUE