/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 52 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 12 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 0 ms]
(6) QTRS
(7) QTRSRRRProof [EQUIVALENT, 0 ms]
(8) QTRS
(9) AAECC Innermost [EQUIVALENT, 0 ms]
(10) QTRS
(11) DependencyPairsProof [EQUIVALENT, 0 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 0 ms]
(14) TRUE


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

   f(0) -> cons(0, n__f(n__s(n__0)))
   f(s(0)) -> f(p(s(0)))
   p(s(0)) -> 0
   f(X) -> n__f(X)
   s(X) -> n__s(X)
   0 -> n__0
   activate(n__f(X)) -> f(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__0) -> 0
   activate(X) -> X

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 1
   POL(activate(x_1)) = 2 + 2*x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(f(x_1)) = 2 + 2*x_1
   POL(n__0) = 0
   POL(n__f(x_1)) = 2 + 2*x_1
   POL(n__s(x_1)) = x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   f(0) -> cons(0, n__f(n__s(n__0)))
   0 -> n__0
   activate(n__0) -> 0
   activate(X) -> X




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

   f(s(0)) -> f(p(s(0)))
   p(s(0)) -> 0
   f(X) -> n__f(X)
   s(X) -> n__s(X)
   activate(n__f(X)) -> f(activate(X))
   activate(n__s(X)) -> s(activate(X))

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 1
   POL(activate(x_1)) = 2 + 2*x_1
   POL(f(x_1)) = 2 + 2*x_1
   POL(n__f(x_1)) = 2 + 2*x_1
   POL(n__s(x_1)) = 1 + x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 1 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   p(s(0)) -> 0
   activate(n__s(X)) -> s(activate(X))




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

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(s(0)) -> f(p(s(0)))
   f(X) -> n__f(X)
   s(X) -> n__s(X)
   activate(n__f(X)) -> f(activate(X))

Q is empty.

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(5) QTRSRRRProof (EQUIVALENT)
Used ordering:
f/1(YES)
s/1)YES(
0/0)
p/1)YES(
n__f/1(YES)
n__s/1)YES(
activate/1(YES)

Quasi precedence:
activate_1 > [f_1, 0] > n__f_1


Status:
f_1: multiset status
0: multiset status
n__f_1: multiset status
activate_1: multiset status

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   f(X) -> n__f(X)
   activate(n__f(X)) -> f(activate(X))




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

(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(s(0)) -> f(p(s(0)))
   s(X) -> n__s(X)

Q is empty.

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

(7) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(f(x_1)) = x_1
   POL(n__s(x_1)) = 2*x_1
   POL(p(x_1)) = x_1
   POL(s(x_1)) = 1 + 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   s(X) -> n__s(X)




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

(8)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(s(0)) -> f(p(s(0)))

Q is empty.

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

The TRS R 2 is 
   f(s(0)) -> f(p(s(0)))

The signature Sigma is {f_1}
----------------------------------------

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

   f(s(0)) -> f(p(s(0)))

The set Q consists of the following terms:

   f(s(0))


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(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:

   F(s(0)) -> F(p(s(0)))

The TRS R consists of the following rules:

   f(s(0)) -> f(p(s(0)))

The set Q consists of the following terms:

   f(s(0))

We have to consider all minimal (P,Q,R)-chains.
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(13) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(14)
TRUE