/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 5 ms]
(4) QTRS
(5) Overlay + Local Confluence [EQUIVALENT, 2 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 1 ms]
(8) QDP
(9) UsableRulesProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QReductionProof [EQUIVALENT, 0 ms]
(12) QDP
(13) QDPSizeChangeProof [EQUIVALENT, 1 ms]
(14) YES


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

   a(a(x)) -> b(b(x))
   b(b(a(x))) -> a(b(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:

   a(a(x)) -> b(b(x))
   a(b(b(x))) -> b(b(a(x)))

Q is empty.

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

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

   a(a(x)) -> b(b(x))




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

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a(b(b(x))) -> b(b(a(x)))

Q is empty.

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(5) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
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(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a(b(b(x))) -> b(b(a(x)))

The set Q consists of the following terms:

   a(b(b(x0)))


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(7) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(b(b(x))) -> A(x)

The TRS R consists of the following rules:

   a(b(b(x))) -> b(b(a(x)))

The set Q consists of the following terms:

   a(b(b(x0)))

We have to consider all minimal (P,Q,R)-chains.
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(9) 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.
----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(b(b(x))) -> A(x)

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

   a(b(b(x0)))

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

(11) 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].

   a(b(b(x0)))


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

   A(b(b(x))) -> A(x)

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

(13) 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:
*A(b(b(x))) -> A(x)
The graph contains the following edges 1 > 1


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(14)
YES