YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 67 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(10) YES


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

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

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

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

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

The TRS R consists of the following rules:

   4(2(4(x1))) -> 4(0(2(5(3(3(5(0(0(2(x1))))))))))
   2(4(2(4(4(x1))))) -> 1(0(2(0(4(1(2(5(0(2(x1))))))))))
   3(4(2(4(5(0(x1)))))) -> 0(0(0(3(5(3(5(5(1(5(x1))))))))))
   3(3(5(4(1(1(x1)))))) -> 1(2(2(1(0(3(1(1(3(1(x1))))))))))
   2(1(3(4(1(3(x1)))))) -> 3(0(0(3(2(4(1(1(0(0(x1))))))))))
   1(4(2(4(2(3(x1)))))) -> 3(3(1(3(5(1(1(1(2(0(x1))))))))))
   2(1(4(0(3(3(x1)))))) -> 1(3(5(0(2(0(2(1(5(3(x1))))))))))
   4(5(0(5(4(1(4(x1))))))) -> 0(1(3(5(0(1(3(5(1(4(x1))))))))))
   2(2(4(5(0(4(4(x1))))))) -> 1(4(5(4(4(4(3(4(0(4(x1))))))))))
   3(4(2(3(5(4(5(x1))))))) -> 4(1(0(5(4(0(5(5(5(2(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 55 less nodes.
----------------------------------------

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

   2^1(4(2(4(4(x1))))) -> 2^1(x1)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   2^1(4(2(4(4(x1))))) -> 2^1(x1)

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

(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:
*2^1(4(2(4(4(x1))))) -> 2^1(x1)
The graph contains the following edges 1 > 1


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(10)
YES