/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 90 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 0 ms]
(4) QTRS
(5) RisEmptyProof [EQUIVALENT, 0 ms]
(6) YES


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

   filter(cons(X), 0, M) -> cons(0)
   filter(cons(X), s(N), M) -> cons(X)
   sieve(cons(0)) -> cons(0)
   sieve(cons(s(N))) -> cons(s(N))
   nats(N) -> cons(N)
   zprimes -> sieve(nats(s(s(0))))

The set Q consists of the following terms:

   filter(cons(x0), 0, x1)
   filter(cons(x0), s(x1), x2)
   sieve(cons(0))
   sieve(cons(s(x0)))
   nats(x0)
   zprimes


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

   POL(0) = 0
   POL(cons(x_1)) = x_1
   POL(filter(x_1, x_2, x_3)) = 1 + 2*x_1 + x_2 + x_3
   POL(nats(x_1)) = 2*x_1
   POL(s(x_1)) = x_1
   POL(sieve(x_1)) = 1 + 2*x_1
   POL(zprimes) = 2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   filter(cons(X), 0, M) -> cons(0)
   filter(cons(X), s(N), M) -> cons(X)
   sieve(cons(0)) -> cons(0)
   sieve(cons(s(N))) -> cons(s(N))
   zprimes -> sieve(nats(s(s(0))))




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

   nats(N) -> cons(N)

The set Q consists of the following terms:

   filter(cons(x0), 0, x1)
   filter(cons(x0), s(x1), x2)
   sieve(cons(0))
   sieve(cons(s(x0)))
   nats(x0)
   zprimes


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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
nats/1(YES)
cons/1)YES(

Quasi precedence:
trivial


Status:
nats_1: multiset status

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

   nats(N) -> cons(N)




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(4)
Obligation:
Q restricted rewrite system:
R is empty.
The set Q consists of the following terms:

   filter(cons(x0), 0, x1)
   filter(cons(x0), s(x1), x2)
   sieve(cons(0))
   sieve(cons(s(x0)))
   nats(x0)
   zprimes


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(5) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(6)
YES