YES(O(1), O(1)) 0.00/0.55 YES(O(1), O(1)) 0.00/0.56 0.00/0.56 0.00/0.56 0.00/0.56 0.00/0.56 0.00/0.56 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.56 0.00/0.56 0.00/0.56
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(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

filter(cons(X), 0, M) → cons(0) 0.00/0.56
filter(cons(X), s(N), M) → cons(X) 0.00/0.56
sieve(cons(0)) → cons(0) 0.00/0.56
sieve(cons(s(N))) → cons(s(N)) 0.00/0.56
nats(N) → cons(N) 0.00/0.56
zprimessieve(nats(s(s(0))))

Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
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(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

filter(cons(z0), 0, z1) → cons(0) 0.00/0.56
filter(cons(z0), s(z1), z2) → cons(z0) 0.00/0.56
sieve(cons(0)) → cons(0) 0.00/0.56
sieve(cons(s(z0))) → cons(s(z0)) 0.00/0.56
nats(z0) → cons(z0) 0.00/0.56
zprimessieve(nats(s(s(0))))
Tuples:

ZPRIMESc5(SIEVE(nats(s(s(0)))), NATS(s(s(0))))
S tuples:

ZPRIMESc5(SIEVE(nats(s(s(0)))), NATS(s(s(0))))
K tuples:none
Defined Rule Symbols:

filter, sieve, nats, zprimes

Defined Pair Symbols:

ZPRIMES

Compound Symbols:

c5

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(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing tuple parts
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

filter(cons(z0), 0, z1) → cons(0) 0.00/0.56
filter(cons(z0), s(z1), z2) → cons(z0) 0.00/0.56
sieve(cons(0)) → cons(0) 0.00/0.56
sieve(cons(s(z0))) → cons(s(z0)) 0.00/0.56
nats(z0) → cons(z0) 0.00/0.56
zprimessieve(nats(s(s(0))))
Tuples:

ZPRIMESc5
S tuples:

ZPRIMESc5
K tuples:none
Defined Rule Symbols:

filter, sieve, nats, zprimes

Defined Pair Symbols:

ZPRIMES

Compound Symbols:

c5

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(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing nodes:

ZPRIMESc5
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0.00/0.56

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

filter(cons(z0), 0, z1) → cons(0) 0.00/0.56
filter(cons(z0), s(z1), z2) → cons(z0) 0.00/0.56
sieve(cons(0)) → cons(0) 0.00/0.56
sieve(cons(s(z0))) → cons(s(z0)) 0.00/0.56
nats(z0) → cons(z0) 0.00/0.56
zprimessieve(nats(s(s(0))))
Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:

filter, sieve, nats, zprimes

Defined Pair Symbols:none

Compound Symbols:none

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(7) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty
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(8) BOUNDS(O(1), O(1))

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0.00/0.58 EOF