/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) MNOCProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) UsableRulesProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QReductionProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES


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

   w(r(x)) -> r(w(x))
   b(r(x)) -> r(b(x))
   b(w(x)) -> w(b(x))

Q is empty.

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(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:

   r(w(x)) -> w(r(x))
   r(b(x)) -> b(r(x))
   w(b(x)) -> b(w(x))

Q is empty.

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

(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:

   R(w(x)) -> W(r(x))
   R(w(x)) -> R(x)
   R(b(x)) -> R(x)
   W(b(x)) -> W(x)

The TRS R consists of the following rules:

   r(w(x)) -> w(r(x))
   r(b(x)) -> b(r(x))
   w(b(x)) -> b(w(x))

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.
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(6)
Complex Obligation (AND)

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

   W(b(x)) -> W(x)

The TRS R consists of the following rules:

   r(w(x)) -> w(r(x))
   r(b(x)) -> b(r(x))
   w(b(x)) -> b(w(x))

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

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

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

   W(b(x)) -> W(x)

The TRS R consists of the following rules:

   r(w(x)) -> w(r(x))
   r(b(x)) -> b(r(x))
   w(b(x)) -> b(w(x))

The set Q consists of the following terms:

   r(w(x0))
   r(b(x0))
   w(b(x0))

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

(10) 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.
----------------------------------------

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

   W(b(x)) -> W(x)

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

   r(w(x0))
   r(b(x0))
   w(b(x0))

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

(12) 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].

   r(w(x0))
   r(b(x0))
   w(b(x0))


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

   W(b(x)) -> W(x)

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

(14) 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:
*W(b(x)) -> W(x)
The graph contains the following edges 1 > 1


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(15)
YES

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

   R(b(x)) -> R(x)
   R(w(x)) -> R(x)

The TRS R consists of the following rules:

   r(w(x)) -> w(r(x))
   r(b(x)) -> b(r(x))
   w(b(x)) -> b(w(x))

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

(17) 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.
----------------------------------------

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

   R(b(x)) -> R(x)
   R(w(x)) -> R(x)

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

(19) 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:
*R(b(x)) -> R(x)
The graph contains the following edges 1 > 1


*R(w(x)) -> R(x)
The graph contains the following edges 1 > 1


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(20)
YES