/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 6 ms]
(2) QDP
(3) MRRProof [EQUIVALENT, 72 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 548 ms]
(8) QDP
(9) PisEmptyProof [EQUIVALENT, 0 ms]
(10) YES


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

   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
   1(0(0(1(x1)))) -> 0(0(0(0(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:

   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))
   0^1(0(0(0(x1)))) -> 1^1(1(1(x1)))
   0^1(0(0(0(x1)))) -> 1^1(1(x1))
   0^1(0(0(0(x1)))) -> 1^1(x1)
   1^1(0(0(1(x1)))) -> 0^1(0(0(0(x1))))
   1^1(0(0(1(x1)))) -> 0^1(0(0(x1)))
   1^1(0(0(1(x1)))) -> 0^1(0(x1))
   1^1(0(0(1(x1)))) -> 0^1(x1)

The TRS R consists of the following rules:

   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
   1(0(0(1(x1)))) -> 0(0(0(0(x1))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) 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:

   0^1(0(0(0(x1)))) -> 1^1(1(x1))
   0^1(0(0(0(x1)))) -> 1^1(x1)
   1^1(0(0(1(x1)))) -> 0^1(0(0(0(x1))))
   1^1(0(0(1(x1)))) -> 0^1(0(0(x1)))
   1^1(0(0(1(x1)))) -> 0^1(0(x1))
   1^1(0(0(1(x1)))) -> 0^1(x1)


Used ordering: Polynomial interpretation [POLO]:

   POL(0(x_1)) = 1 + x_1
   POL(0^1(x_1)) = 2*x_1
   POL(1(x_1)) = 1 + x_1
   POL(1^1(x_1)) = 2 + 2*x_1


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

   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))
   0^1(0(0(0(x1)))) -> 1^1(1(1(x1)))

The TRS R consists of the following rules:

   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
   1(0(0(1(x1)))) -> 0(0(0(0(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 1 SCC with 1 less node.
----------------------------------------

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

   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))

The TRS R consists of the following rules:

   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
   1(0(0(1(x1)))) -> 0(0(0(0(x1))))

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

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


The following pairs can be oriented strictly and are deleted.

   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

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

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

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


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

   1(0(0(1(x1)))) -> 0(0(0(0(x1))))
   0(0(0(0(x1)))) -> 0(1(1(1(x1))))


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

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

   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
   1(0(0(1(x1)))) -> 0(0(0(0(x1))))

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

(9) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(10)
YES