/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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 w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 113 ms]
(2) QTRS
(3) AAECC Innermost [EQUIVALENT, 0 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) 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(X)) -> X
   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:
f/1(YES)
0/0)
cons/2(YES,YES)
n__f/1(YES)
n__s/1(YES)
n__0/0)
s/1(YES)
p/1)YES(
activate/1(YES)

Quasi precedence:
activate_1 > s_1 > f_1 > [0, n__s_1] > cons_2
activate_1 > s_1 > f_1 > [0, n__s_1] > n__f_1
activate_1 > s_1 > f_1 > [0, n__s_1] > n__0


Status:
f_1: multiset status
0: multiset status
cons_2: multiset status
n__f_1: multiset status
n__s_1: multiset status
n__0: multiset status
s_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(0) -> cons(0, n__f(n__s(n__0)))
   p(s(X)) -> X
   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




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

(2)
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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(3) 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}
----------------------------------------

(4)
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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(5) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(6)
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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(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(8)
TRUE