/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 11 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 6 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 175 ms]
(6) QDP
(7) MRRProof [EQUIVALENT, 43 ms]
(8) QDP
(9) QDPOrderProof [EQUIVALENT, 60 ms]
(10) QDP
(11) QDPOrderProof [EQUIVALENT, 279 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 0 ms]
(14) TRUE


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

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

   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   a(a(x1)) -> b(b(b(b(b(b(x1))))))
   b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))
   c(c(x1)) -> d(d(x1))

Q is empty.

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

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

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(a(d(d(x1)))) -> B(b(x1))
   A(a(d(d(x1)))) -> B(x1)
   A(a(x1)) -> B(b(b(b(b(b(x1))))))
   A(a(x1)) -> B(b(b(b(b(x1)))))
   A(a(x1)) -> B(b(b(b(x1))))
   A(a(x1)) -> B(b(b(x1)))
   A(a(x1)) -> B(b(x1))
   A(a(x1)) -> B(x1)
   B(b(d(d(b(b(x1)))))) -> A(a(c(c(x1))))
   B(b(d(d(b(b(x1)))))) -> A(c(c(x1)))
   B(b(d(d(b(b(x1)))))) -> C(c(x1))
   B(b(d(d(b(b(x1)))))) -> C(x1)

The TRS R consists of the following rules:

   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   a(a(x1)) -> b(b(b(b(b(b(x1))))))
   b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))
   c(c(x1)) -> d(d(x1))

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

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

   B(b(d(d(b(b(x1)))))) -> A(a(c(c(x1))))
   A(a(d(d(x1)))) -> B(b(x1))
   B(b(d(d(b(b(x1)))))) -> A(c(c(x1)))
   A(a(d(d(x1)))) -> B(x1)
   A(a(x1)) -> B(b(b(b(b(b(x1))))))
   A(a(x1)) -> B(b(b(b(b(x1)))))
   A(a(x1)) -> B(b(b(b(x1))))
   A(a(x1)) -> B(b(b(x1)))
   A(a(x1)) -> B(b(x1))
   A(a(x1)) -> B(x1)

The TRS R consists of the following rules:

   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   a(a(x1)) -> b(b(b(b(b(b(x1))))))
   b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))
   c(c(x1)) -> d(d(x1))

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

(5) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   B(b(d(d(b(b(x1)))))) -> A(c(c(x1)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2*x_1
   POL(B(x_1)) = 2
   POL(a(x_1)) = 1
   POL(b(x_1)) = 1
   POL(c(x_1)) = 0
   POL(d(x_1)) = 0

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

   c(c(x1)) -> d(d(x1))
   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   a(a(x1)) -> b(b(b(b(b(b(x1))))))
   b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))


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

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

   B(b(d(d(b(b(x1)))))) -> A(a(c(c(x1))))
   A(a(d(d(x1)))) -> B(b(x1))
   A(a(d(d(x1)))) -> B(x1)
   A(a(x1)) -> B(b(b(b(b(b(x1))))))
   A(a(x1)) -> B(b(b(b(b(x1)))))
   A(a(x1)) -> B(b(b(b(x1))))
   A(a(x1)) -> B(b(b(x1)))
   A(a(x1)) -> B(b(x1))
   A(a(x1)) -> B(x1)

The TRS R consists of the following rules:

   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   a(a(x1)) -> b(b(b(b(b(b(x1))))))
   b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))
   c(c(x1)) -> d(d(x1))

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

(7) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   A(a(d(d(x1)))) -> B(b(x1))
   A(a(d(d(x1)))) -> B(x1)


Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2*x_1
   POL(B(x_1)) = 2*x_1
   POL(a(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = 2 + 2*x_1
   POL(d(x_1)) = 2 + 2*x_1


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

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

   B(b(d(d(b(b(x1)))))) -> A(a(c(c(x1))))
   A(a(x1)) -> B(b(b(b(b(b(x1))))))
   A(a(x1)) -> B(b(b(b(b(x1)))))
   A(a(x1)) -> B(b(b(b(x1))))
   A(a(x1)) -> B(b(b(x1)))
   A(a(x1)) -> B(b(x1))
   A(a(x1)) -> B(x1)

The TRS R consists of the following rules:

   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   a(a(x1)) -> b(b(b(b(b(b(x1))))))
   b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))
   c(c(x1)) -> d(d(x1))

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

(9) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   A(a(x1)) -> B(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(A(x_1)) = 4*x_1
   POL(B(x_1)) = 2*x_1
   POL(a(x_1)) = 2 + x_1
   POL(b(x_1)) = 4
   POL(c(x_1)) = 0
   POL(d(x_1)) = 0

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

   c(c(x1)) -> d(d(x1))
   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   a(a(x1)) -> b(b(b(b(b(b(x1))))))
   b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))


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

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

   B(b(d(d(b(b(x1)))))) -> A(a(c(c(x1))))
   A(a(x1)) -> B(b(b(b(b(b(x1))))))
   A(a(x1)) -> B(b(b(b(b(x1)))))
   A(a(x1)) -> B(b(b(b(x1))))
   A(a(x1)) -> B(b(b(x1)))
   A(a(x1)) -> B(b(x1))

The TRS R consists of the following rules:

   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   a(a(x1)) -> b(b(b(b(b(b(x1))))))
   b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))
   c(c(x1)) -> d(d(x1))

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

(11) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   B(b(d(d(b(b(x1)))))) -> A(a(c(c(x1))))
   A(a(x1)) -> B(b(b(b(b(x1)))))
   A(a(x1)) -> B(b(b(b(x1))))
   A(a(x1)) -> B(b(b(x1)))
   A(a(x1)) -> B(b(x1))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2*x_1
   POL(B(x_1)) = 2*x_1
   POL(a(x_1)) = 5 + x_1
   POL(b(x_1)) = 1 + x_1
   POL(c(x_1)) = 2*x_1
   POL(d(x_1)) = 2*x_1

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

   c(c(x1)) -> d(d(x1))
   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   a(a(x1)) -> b(b(b(b(b(b(x1))))))
   b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))


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

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(a(x1)) -> B(b(b(b(b(b(x1))))))

The TRS R consists of the following rules:

   a(a(d(d(x1)))) -> d(d(b(b(x1))))
   a(a(x1)) -> b(b(b(b(b(b(x1))))))
   b(b(d(d(b(b(x1)))))) -> a(a(c(c(x1))))
   c(c(x1)) -> d(d(x1))

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

(13) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(14)
TRUE