/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination of the given CSR could be proven:

(0) CSR
(1) CSRRRRProof [EQUIVALENT, 35 ms]
(2) CSR
(3) CSDependencyPairsProof [EQUIVALENT, 0 ms]
(4) QCSDP
(5) QCSDependencyGraphProof [EQUIVALENT, 0 ms]
(6) TRUE


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(0)
Obligation:
Context-sensitive rewrite system:
The TRS R consists of the following rules:

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

The replacement map contains the following entries:

f: {1}
a: empty set
b: empty set

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(1) CSRRRRProof (EQUIVALENT)
The following CSR is given: Context-sensitive rewrite system:
The TRS R consists of the following rules:

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

The replacement map contains the following entries:

f: {1}
a: empty set
b: empty set
Used ordering:
Polynomial interpretation [POLO]:

   POL(a) = 0
   POL(b) = 2
   POL(f(x_1, x_2)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   b -> a




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

(2)
Obligation:
Context-sensitive rewrite system:
The TRS R consists of the following rules:

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

The replacement map contains the following entries:

f: {1}
a: empty set
b: empty set

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

(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_2, F_2} we have mu(f) = {1}.

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

The TRS R consists of the following rules:

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

Q is empty.

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(5) QCSDependencyGraphProof (EQUIVALENT)
The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 0 SCCs.
The rules F(z0, z0) -> F(a, b) and F(x0, x0) -> F(a, b) form no chain, because ECap^mu(F(a, b)) = F(a, b) does not unify with F(x0, x0). 
----------------------------------------

(6)
TRUE