/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 7 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 32 ms]
(8) QDP
(9) PisEmptyProof [EQUIVALENT, 0 ms]
(10) YES


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

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

   a(a(f(b, a(x)))) -> f(a(a(a(x))), b)
   a(a(x)) -> f(b, a(f(a(x), b)))
   f(a(x), b) -> f(b, a(x))

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(f(b, a(x)))) -> F(a(a(a(x))), b)
   A(a(f(b, a(x)))) -> A(a(a(x)))
   A(a(f(b, a(x)))) -> A(a(x))
   A(a(x)) -> F(b, a(f(a(x), b)))
   A(a(x)) -> A(f(a(x), b))
   A(a(x)) -> F(a(x), b)
   F(a(x), b) -> F(b, a(x))

The TRS R consists of the following rules:

   a(a(f(b, a(x)))) -> f(a(a(a(x))), b)
   a(a(x)) -> f(b, a(f(a(x), b)))
   f(a(x), b) -> f(b, a(x))

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 5 less nodes.
----------------------------------------

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

   A(a(f(b, a(x)))) -> A(a(x))
   A(a(f(b, a(x)))) -> A(a(a(x)))

The TRS R consists of the following rules:

   a(a(f(b, a(x)))) -> f(a(a(a(x))), b)
   a(a(x)) -> f(b, a(f(a(x), b)))
   f(a(x), b) -> f(b, a(x))

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

(5) 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(f(b, a(x)))) -> A(a(x))


Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2*x_1
   POL(a(x_1)) = 1 + 2*x_1
   POL(b) = 0
   POL(f(x_1, x_2)) = x_1 + x_2


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

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

   A(a(f(b, a(x)))) -> A(a(a(x)))

The TRS R consists of the following rules:

   a(a(f(b, a(x)))) -> f(a(a(a(x))), b)
   a(a(x)) -> f(b, a(f(a(x), b)))
   f(a(x), b) -> f(b, a(x))

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.

   A(a(f(b, a(x)))) -> A(a(a(x)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

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

   <<<
 POL(a(x_1)) =  	[[0], [1]] 	 +  	[[0, 1], [1, 0]] 	* 	x_1
>>>

   <<<
 POL(f(x_1, x_2)) =  	[[1], [0]] 	 +  	[[1, 0], [0, 0]] 	* 	x_1 	 +  	[[1, 0], [0, 0]] 	* 	x_2
>>>

   <<<
 POL(b) =  	[[0], [0]]
>>>


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

   a(a(f(b, a(x)))) -> f(a(a(a(x))), b)
   a(a(x)) -> f(b, a(f(a(x), b)))
   f(a(x), b) -> f(b, a(x))


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

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

   a(a(f(b, a(x)))) -> f(a(a(a(x))), b)
   a(a(x)) -> f(b, a(f(a(x), b)))
   f(a(x), b) -> f(b, a(x))

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.
----------------------------------------

(10)
YES