/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 62 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) MRRProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPOrderProof [EQUIVALENT, 37 ms]
(10) QDP
(11) PisEmptyProof [EQUIVALENT, 0 ms]
(12) YES


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

   a(a(y, 0), 0) -> y
   c(c(y)) -> y
   c(a(c(c(y)), x)) -> a(c(c(c(a(x, 0)))), y)

Q is empty.

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

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

   POL(0) = 0
   POL(a(x_1, x_2)) = x_1 + x_2
   POL(c(x_1)) = 2 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   c(c(y)) -> y




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

   a(a(y, 0), 0) -> y
   c(a(c(c(y)), x)) -> a(c(c(c(a(x, 0)))), y)

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:

   C(a(c(c(y)), x)) -> A(c(c(c(a(x, 0)))), y)
   C(a(c(c(y)), x)) -> C(c(c(a(x, 0))))
   C(a(c(c(y)), x)) -> C(c(a(x, 0)))
   C(a(c(c(y)), x)) -> C(a(x, 0))
   C(a(c(c(y)), x)) -> A(x, 0)

The TRS R consists of the following rules:

   a(a(y, 0), 0) -> y
   c(a(c(c(y)), x)) -> a(c(c(c(a(x, 0)))), y)

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 2 less nodes.
----------------------------------------

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

   C(a(c(c(y)), x)) -> C(c(a(x, 0)))
   C(a(c(c(y)), x)) -> C(c(c(a(x, 0))))
   C(a(c(c(y)), x)) -> C(a(x, 0))

The TRS R consists of the following rules:

   a(a(y, 0), 0) -> y
   c(a(c(c(y)), x)) -> a(c(c(c(a(x, 0)))), y)

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:

   C(a(c(c(y)), x)) -> C(c(a(x, 0)))
   C(a(c(c(y)), x)) -> C(a(x, 0))


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(C(x_1)) = x_1
   POL(a(x_1, x_2)) = x_1 + x_2
   POL(c(x_1)) = 1 + x_1


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

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

   C(a(c(c(y)), x)) -> C(c(c(a(x, 0))))

The TRS R consists of the following rules:

   a(a(y, 0), 0) -> y
   c(a(c(c(y)), x)) -> a(c(c(c(a(x, 0)))), y)

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.

   C(a(c(c(y)), x)) -> C(c(c(a(x, 0))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

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

   <<<
 POL(a(x_1, x_2)) =  	[[-I]] 	 +  	[[1A]] 	* 	x_1 	 +  	[[4A]] 	* 	x_2
>>>

   <<<
 POL(c(x_1)) =  	[[5A]] 	 +  	[[1A]] 	* 	x_1
>>>

   <<<
 POL(0) =  	[[0A]]
>>>


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

   a(a(y, 0), 0) -> y
   c(a(c(c(y)), x)) -> a(c(c(c(a(x, 0)))), y)


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

   a(a(y, 0), 0) -> y
   c(a(c(c(y)), x)) -> a(c(c(c(a(x, 0)))), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(11) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(12)
YES