/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) TransformationProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPOrderProof [EQUIVALENT, 19.7 s]
        (18) QDP
        (19) PisEmptyProof [EQUIVALENT, 0 ms]
        (20) YES


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

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

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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:

   P(a(x0), p(b(a(x1)), x2)) -> P(x1, p(a(b(a(x1))), x2))
   P(a(x0), p(b(a(x1)), x2)) -> P(a(b(a(x1))), x2)
   P(a(x0), p(b(a(x1)), x2)) -> A(b(a(x1)))
   A(b(a(x0))) -> A(b(x0))

The TRS R consists of the following rules:

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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 2 SCCs with 1 less node.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(b(a(x0))) -> A(b(x0))

The TRS R consists of the following rules:

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(b(a(x0))) -> A(b(x0))

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

(8) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

   A(b(a(x0))) -> A(b(x0))
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2*x_1
   POL(a(x_1)) = x_1
   POL(b(x_1)) = 2*x_1


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

(9)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(11)
YES

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

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

   P(a(x0), p(b(a(x1)), x2)) -> P(a(b(a(x1))), x2)
   P(a(x0), p(b(a(x1)), x2)) -> P(x1, p(a(b(a(x1))), x2))

The TRS R consists of the following rules:

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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

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


The following pairs can be oriented strictly and are deleted.

   P(a(x0), p(b(a(x1)), x2)) -> P(a(b(a(x1))), x2)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
P(x1, x2)  =  x2

p(x1, x2)  =  p(x2)


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   p_1=1

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

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))


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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   P(a(x0), p(b(a(x1)), x2)) -> P(x1, p(a(b(a(x1))), x2))

The TRS R consists of the following rules:

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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

(15) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule P(a(x0), p(b(a(x1)), x2)) -> P(x1, p(a(b(a(x1))), x2)) we obtained the following new rules [LPAR04]:

   (P(a(x0), p(b(a(a(y_0))), x2)) -> P(a(y_0), p(a(b(a(a(y_0)))), x2)),P(a(x0), p(b(a(a(y_0))), x2)) -> P(a(y_0), p(a(b(a(a(y_0)))), x2)))


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

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   P(a(x0), p(b(a(a(y_0))), x2)) -> P(a(y_0), p(a(b(a(a(y_0)))), x2))

The TRS R consists of the following rules:

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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

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


The following pairs can be oriented strictly and are deleted.

   P(a(x0), p(b(a(a(y_0))), x2)) -> P(a(y_0), p(a(b(a(a(y_0)))), x2))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] to (N^4, +, *, >=, >) :

   <<<
 POL(P(x_1, x_2)) =  	[[0]] 	 +  	[[0, 0, 0, 0]] 	* 	x_1 	 +  	[[0, 0, 0, 2]] 	* 	x_2
>>>

   <<<
 POL(a(x_1)) =  	[[0], [0], [1], [0]] 	 +  	[[0, 1, 0, 0], [0, 0, 0, 1], [0, 0, 0, 0], [1, 0, 2, 2]] 	* 	x_1
>>>

   <<<
 POL(p(x_1, x_2)) =  	[[0], [0], [0], [0]] 	 +  	[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [2, 0, 0, 0]] 	* 	x_1 	 +  	[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 1]] 	* 	x_2
>>>

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


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

   a(b(a(x0))) -> b(a(b(x0)))
   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))


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

(18)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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

(19) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(20)
YES