/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 27 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) MNOCProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(8) YES


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

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(i, 1) -> 1
   app(i, app(i, x)) -> x
   app(app(map, f), nil) -> nil
   app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
   app(app(filter, f), nil) -> nil
   app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs)
   app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs))
   app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs)

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))
   APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x))
   APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(filter2, app(f, x)), f), x)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter2, app(f, x)), f)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x))
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs))
   APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f)

The TRS R consists of the following rules:

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(i, 1) -> 1
   app(i, app(i, x)) -> x
   app(app(map, f), nil) -> nil
   app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
   app(app(filter, f), nil) -> nil
   app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs)
   app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs))
   app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 9 less nodes.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)

The TRS R consists of the following rules:

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(i, 1) -> 1
   app(i, app(i, x)) -> x
   app(app(map, f), nil) -> nil
   app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
   app(app(filter, f), nil) -> nil
   app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs)
   app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs))
   app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)

The TRS R consists of the following rules:

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(i, 1) -> 1
   app(i, app(i, x)) -> x
   app(app(map, f), nil) -> nil
   app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
   app(app(filter, f), nil) -> nil
   app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs)
   app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs))
   app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs)

The set Q consists of the following terms:

   app(app(., 1), x0)
   app(app(., x0), 1)
   app(app(., app(i, x0)), x0)
   app(app(., x0), app(i, x0))
   app(app(., app(i, x0)), app(app(., x0), x1))
   app(app(., x0), app(app(., app(i, x0)), x1))
   app(i, 1)
   app(i, app(i, x0))
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))
   app(app(filter, x0), nil)
   app(app(filter, x0), app(app(cons, x1), x2))
   app(app(app(app(filter2, true), x0), x1), x2)
   app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
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(7) 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:
*APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
The graph contains the following edges 1 > 1, 2 > 2


*APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
The graph contains the following edges 1 > 1, 2 > 2


*APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
The graph contains the following edges 1 >= 1, 2 > 2


*APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs)
The graph contains the following edges 2 > 2


*APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
The graph contains the following edges 2 >= 2


*APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)
The graph contains the following edges 2 >= 2


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(8)
YES