/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSToCSRProof [EQUIVALENT, 0 ms]
(2) CSR
(3) CSDependencyPairsProof [EQUIVALENT, 0 ms]
(4) QCSDP
(5) QCSDependencyGraphProof [EQUIVALENT, 0 ms]
(6) QCSDP
(7) QCSUsableRulesProof [EQUIVALENT, 0 ms]
(8) QCSDP
(9) QCSDPInstantiationProcessor [EQUIVALENT, 0 ms]
(10) QCSDP
(11) PIsEmptyProof [EQUIVALENT, 0 ms]
(12) YES


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

   active(f(a, X, X)) -> mark(f(X, b, b))
   active(b) -> mark(a)
   active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
   f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
   proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
   proper(a) -> ok(a)
   proper(b) -> ok(b)
   f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

Q is empty.

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(1) QTRSToCSRProof (EQUIVALENT)
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

   active(f(a, X, X)) -> mark(f(X, b, b))
   active(b) -> mark(a)
   active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
   f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
   proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
   proper(a) -> ok(a)
   proper(b) -> ok(b)
   f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
   top(mark(X)) -> top(proper(X))
   top(ok(X)) -> top(active(X))

Q is empty.
Special symbols used for the transformation (see [GM04]):
top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1
The replacement map contains the following entries:

f: {2}
a: empty set
b: empty set
The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound).
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(2)
Obligation:
Context-sensitive rewrite system:
The TRS R consists of the following rules:

   f(a, X, X) -> f(X, b, b)
   b -> a

The replacement map contains the following entries:

f: {2}
a: empty set
b: empty set

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(3) CSDependencyPairsProof (EQUIVALENT)
Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.
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(4)
Obligation:
Q-restricted context-sensitive dependency pair problem:
For all symbols f in {f_3, F_3} we have mu(f) = {2}.

The ordinary context-sensitive dependency pairs DP_o are:
   F(a, X, X) -> F(X, b, b)
   F(a, X, X) -> B

The TRS R consists of the following rules:

   f(a, X, X) -> f(X, b, b)
   b -> a

Q is empty.

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(5) QCSDependencyGraphProof (EQUIVALENT)
The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 1 SCC with 1 less node.

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(6)
Obligation:
Q-restricted context-sensitive dependency pair problem:
For all symbols f in {f_3, F_3} we have mu(f) = {2}.

The TRS P consists of the following rules:

   F(a, X, X) -> F(X, b, b)

The TRS R consists of the following rules:

   f(a, X, X) -> f(X, b, b)
   b -> a

Q is empty.

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(7) QCSUsableRulesProof (EQUIVALENT)
The following rules are not useable [DA_EMMES] and can be deleted:

   f(a, x0, x0) -> f(x0, b, b)

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(8)
Obligation:
Q-restricted context-sensitive dependency pair problem:
For all symbols f in {F_3} we have mu(f) = {2}.

The TRS P consists of the following rules:

   F(a, X, X) -> F(X, b, b)

The TRS R consists of the following rules:

   b -> a

Q is empty.

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(9) QCSDPInstantiationProcessor (EQUIVALENT)
Using the Context-Sensitive Instantiation[LPAR08,DA_EMMES] Processor

the pair F(a, X, X) -> F(X, b, b)

was transformed to the following new pairs: 
   F(a, b, b) -> F(b, b, b)



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

(10)
Obligation:
Q-restricted context-sensitive dependency pair problem:

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

   b -> a

Q is empty.

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(11) PIsEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R,mu)-chain.
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(12)
YES