/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 25 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 1254 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 10 ms]
(6) QDP
(7) PisEmptyProof [EQUIVALENT, 0 ms]
(8) YES


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

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

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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


The following pairs can be oriented strictly and are deleted.

   B(b(c(c(a(a(x1)))))) -> A(a(a(a(b(b(x1))))))
   B(b(c(c(a(a(x1)))))) -> A(a(a(b(b(x1)))))
   B(b(c(c(a(a(x1)))))) -> A(a(b(b(x1))))
   B(b(c(c(a(a(x1)))))) -> A(b(b(x1)))
   B(b(c(c(a(a(x1)))))) -> B(b(x1))
   B(b(c(c(a(a(x1)))))) -> B(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(A(x_1)) =  	[[0A]] 	 +  	[[0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(a(x_1)) =  	[[-I], [0A], [-I]] 	 +  	[[0A, -I, 0A], [0A, 0A, -I], [0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(b(x_1)) =  	[[0A], [0A], [0A]] 	 +  	[[0A, 0A, 0A], [0A, -I, -I], [0A, 0A, -I]] 	* 	x_1
>>>

   <<<
 POL(B(x_1)) =  	[[0A]] 	 +  	[[0A, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(c(x_1)) =  	[[0A], [-I], [0A]] 	 +  	[[-I, -I, -I], [1A, -I, -I], [0A, 0A, 0A]] 	* 	x_1
>>>


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

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


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

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

The TRS R consists of the following rules:

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

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


The following pairs can be oriented strictly and are deleted.

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

   POL(A(x_1)) = 1 + x_1
   POL(B(x_1)) = 1 + x_1
   POL(a(x_1)) = 1 + x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = 0

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

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


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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(8)
YES