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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 13 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 3 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) DependencyGraphProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) MNOCProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) UsableRulesProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) QReductionProof [EQUIVALENT, 1 ms]
        (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) MRRProof [EQUIVALENT, 23 ms]
        (22) QDP
        (23) QDPOrderProof [EQUIVALENT, 717 ms]
        (24) QDP
        (25) PisEmptyProof [EQUIVALENT, 0 ms]
        (26) YES


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

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

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

Q is empty.

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

(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

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

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

Q is empty.

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

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

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

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

The TRS R consists of the following rules:

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

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 2 SCCs with 3 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

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

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

(8) 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.
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

(10) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(a(a(a(x1)))) -> B(x1)

The TRS R consists of the following rules:

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

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

(12) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(a(a(a(x1)))) -> B(x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   b(a(a(a(x0))))

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

(14) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(a(a(a(x1)))) -> B(x1)

R is empty.
The set Q consists of the following terms:

   b(a(a(a(x0))))

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

(16) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   b(a(a(a(x0))))


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

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(a(a(a(x1)))) -> B(x1)

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

(18) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*B(a(a(a(x1)))) -> B(x1)
The graph contains the following edges 1 > 1


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

(19)
YES

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

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


Used ordering: Polynomial interpretation [POLO]:

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


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

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

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

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

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

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


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

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


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

(24)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

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

(26)
YES