/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 30 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 9 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 5 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 1 ms]
        (9) QDP
        (10) DependencyGraphProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) UsableRulesProof [EQUIVALENT, 1 ms]
        (13) QDP
        (14) MRRProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) PisEmptyProof [EQUIVALENT, 0 ms]
        (17) YES
    (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 1 ms]
        (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (22) YES


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

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

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

Q is empty.

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

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

   POL(a(x_1)) = x_1
   POL(b(x_1)) = 1 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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




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

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

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

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

The TRS R consists of the following rules:

   b(b(a(a(x1)))) -> a(b(a(b(x1))))
   b(a(a(b(x1)))) -> b(a(b(a(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 1 less node.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

   b(a(a(b(x1)))) -> b(a(b(a(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 1 less node.
----------------------------------------

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

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

The TRS R consists of the following rules:

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

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

(12) 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.
----------------------------------------

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

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

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

(14) 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:

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


Used ordering: Polynomial interpretation [POLO]:

   POL(B(x_1)) = x_1
   POL(a(x_1)) = 2 + x_1
   POL(b(x_1)) = 2*x_1


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

(15)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(17)
YES

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

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

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

The TRS R consists of the following rules:

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

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

(19) 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.
----------------------------------------

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

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

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

(21) 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(b(a(a(x1)))) -> B(x1)
The graph contains the following edges 1 > 1


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(22)
YES