/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) UsableRulesProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QReductionProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPToSRSProof [SOUND, 0 ms]
(10) QTRS
(11) DependencyPairsProof [EQUIVALENT, 0 ms]
(12) QDP
(13) MNOCProof [EQUIVALENT, 0 ms]
(14) QDP
(15) UsableRulesProof [EQUIVALENT, 0 ms]
(16) QDP
(17) QReductionProof [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:

   f(x, a(b(y))) -> f(a(b(b(x))), y)
   f(a(x), y) -> f(x, a(y))
   f(b(x), y) -> f(x, b(y))

Q is empty.

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(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is 
   f(x, a(b(y))) -> f(a(b(b(x))), y)
   f(a(x), y) -> f(x, a(y))
   f(b(x), y) -> f(x, b(y))

The signature Sigma is {f_2}
----------------------------------------

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

   f(x, a(b(y))) -> f(a(b(b(x))), y)
   f(a(x), y) -> f(x, a(y))
   f(b(x), y) -> f(x, b(y))

The set Q consists of the following terms:

   f(x0, a(b(x1)))
   f(a(x0), x1)
   f(b(x0), x1)


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(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:

   F(x, a(b(y))) -> F(a(b(b(x))), y)
   F(a(x), y) -> F(x, a(y))
   F(b(x), y) -> F(x, b(y))

The TRS R consists of the following rules:

   f(x, a(b(y))) -> f(a(b(b(x))), y)
   f(a(x), y) -> f(x, a(y))
   f(b(x), y) -> f(x, b(y))

The set Q consists of the following terms:

   f(x0, a(b(x1)))
   f(a(x0), x1)
   f(b(x0), x1)

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

(5) 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.
----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, a(b(y))) -> F(a(b(b(x))), y)
   F(a(x), y) -> F(x, a(y))
   F(b(x), y) -> F(x, b(y))

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

   f(x0, a(b(x1)))
   f(a(x0), x1)
   f(b(x0), x1)

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

(7) 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].

   f(x0, a(b(x1)))
   f(a(x0), x1)
   f(b(x0), x1)


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

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, a(b(y))) -> F(a(b(b(x))), y)
   F(a(x), y) -> F(x, a(y))
   F(b(x), y) -> F(x, b(y))

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

(9) QDPToSRSProof (SOUND)
The finiteness of this DP problem is implied by strong termination of a SRS due to [UNKNOWN].

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

(10)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

Q is empty.

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   a(b(x0))

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

(15) 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.
----------------------------------------

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

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

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

   a(b(x0))

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

(17) 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(x0))


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

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


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(20)
YES