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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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


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

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

   sort(nil) -> nil
   sort(cons(x, y)) -> insert(x, sort(y))
   insert(x, nil) -> cons(x, nil)
   insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
   choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
   choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
   choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)

Q is empty.

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

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

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

   sort(nil) -> nil
   sort(cons(x, y)) -> insert(x, sort(y))
   insert(x, nil) -> cons(x, nil)
   insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
   choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
   choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
   choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)

The set Q consists of the following terms:

   sort(nil)
   sort(cons(x0, x1))
   insert(x0, nil)
   insert(x0, cons(x1, x2))
   choose(x0, cons(x1, x2), x3, 0)
   choose(x0, cons(x1, x2), 0, s(x3))
   choose(x0, cons(x1, x2), s(x3), s(x4))


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

(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:

   SORT(cons(x, y)) -> INSERT(x, sort(y))
   SORT(cons(x, y)) -> SORT(y)
   INSERT(x, cons(v, w)) -> CHOOSE(x, cons(v, w), x, v)
   CHOOSE(x, cons(v, w), 0, s(z)) -> INSERT(x, w)
   CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)

The TRS R consists of the following rules:

   sort(nil) -> nil
   sort(cons(x, y)) -> insert(x, sort(y))
   insert(x, nil) -> cons(x, nil)
   insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
   choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
   choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
   choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)

The set Q consists of the following terms:

   sort(nil)
   sort(cons(x0, x1))
   insert(x0, nil)
   insert(x0, cons(x1, x2))
   choose(x0, cons(x1, x2), x3, 0)
   choose(x0, cons(x1, x2), 0, s(x3))
   choose(x0, cons(x1, x2), s(x3), s(x4))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
----------------------------------------

(6)
Complex Obligation (AND)

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

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CHOOSE(x, cons(v, w), 0, s(z)) -> INSERT(x, w)
   INSERT(x, cons(v, w)) -> CHOOSE(x, cons(v, w), x, v)
   CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)

The TRS R consists of the following rules:

   sort(nil) -> nil
   sort(cons(x, y)) -> insert(x, sort(y))
   insert(x, nil) -> cons(x, nil)
   insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
   choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
   choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
   choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)

The set Q consists of the following terms:

   sort(nil)
   sort(cons(x0, x1))
   insert(x0, nil)
   insert(x0, cons(x1, x2))
   choose(x0, cons(x1, x2), x3, 0)
   choose(x0, cons(x1, x2), 0, s(x3))
   choose(x0, cons(x1, x2), s(x3), s(x4))

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

(8) 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.
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CHOOSE(x, cons(v, w), 0, s(z)) -> INSERT(x, w)
   INSERT(x, cons(v, w)) -> CHOOSE(x, cons(v, w), x, v)
   CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)

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

   sort(nil)
   sort(cons(x0, x1))
   insert(x0, nil)
   insert(x0, cons(x1, x2))
   choose(x0, cons(x1, x2), x3, 0)
   choose(x0, cons(x1, x2), 0, s(x3))
   choose(x0, cons(x1, x2), s(x3), s(x4))

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

(10) 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].

   sort(nil)
   sort(cons(x0, x1))
   insert(x0, nil)
   insert(x0, cons(x1, x2))
   choose(x0, cons(x1, x2), x3, 0)
   choose(x0, cons(x1, x2), 0, s(x3))
   choose(x0, cons(x1, x2), s(x3), s(x4))


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CHOOSE(x, cons(v, w), 0, s(z)) -> INSERT(x, w)
   INSERT(x, cons(v, w)) -> CHOOSE(x, cons(v, w), x, v)
   CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)

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

(12) 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:
*INSERT(x, cons(v, w)) -> CHOOSE(x, cons(v, w), x, v)
The graph contains the following edges 1 >= 1, 2 >= 2, 1 >= 3, 2 > 4


*CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3, 4 > 4


*CHOOSE(x, cons(v, w), 0, s(z)) -> INSERT(x, w)
The graph contains the following edges 1 >= 1, 2 > 2


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(13)
YES

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

   SORT(cons(x, y)) -> SORT(y)

The TRS R consists of the following rules:

   sort(nil) -> nil
   sort(cons(x, y)) -> insert(x, sort(y))
   insert(x, nil) -> cons(x, nil)
   insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
   choose(x, cons(v, w), y, 0) -> cons(x, cons(v, w))
   choose(x, cons(v, w), 0, s(z)) -> cons(v, insert(x, w))
   choose(x, cons(v, w), s(y), s(z)) -> choose(x, cons(v, w), y, z)

The set Q consists of the following terms:

   sort(nil)
   sort(cons(x0, x1))
   insert(x0, nil)
   insert(x0, cons(x1, x2))
   choose(x0, cons(x1, x2), x3, 0)
   choose(x0, cons(x1, x2), 0, s(x3))
   choose(x0, cons(x1, x2), s(x3), s(x4))

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:

   SORT(cons(x, y)) -> SORT(y)

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

   sort(nil)
   sort(cons(x0, x1))
   insert(x0, nil)
   insert(x0, cons(x1, x2))
   choose(x0, cons(x1, x2), x3, 0)
   choose(x0, cons(x1, x2), 0, s(x3))
   choose(x0, cons(x1, x2), s(x3), s(x4))

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].

   sort(nil)
   sort(cons(x0, x1))
   insert(x0, nil)
   insert(x0, cons(x1, x2))
   choose(x0, cons(x1, x2), x3, 0)
   choose(x0, cons(x1, x2), 0, s(x3))
   choose(x0, cons(x1, x2), s(x3), s(x4))


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

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

   SORT(cons(x, y)) -> SORT(y)

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:
*SORT(cons(x, y)) -> SORT(y)
The graph contains the following edges 1 > 1


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(20)
YES