YES(O(1), O(n^1)) 0.00/0.79 YES(O(1), O(n^1)) 0.00/0.81 0.00/0.81 0.00/0.81 0.00/0.81 0.00/0.81 0.00/0.81 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.81 0.00/0.81 0.00/0.81
0.00/0.81 0.00/0.81 0.00/0.81
0.00/0.81
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(0) Obligation:

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

f(0) → cons(0, n__f(s(0))) 0.00/0.81
f(s(0)) → f(p(s(0))) 0.00/0.81
p(s(0)) → 0 0.00/0.81
f(X) → n__f(X) 0.00/0.81
activate(n__f(X)) → f(X) 0.00/0.81
activate(X) → X

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

Converted CpxTRS to CDT
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0.00/0.81

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0))) 0.00/0.81
f(s(0)) → f(p(s(0))) 0.00/0.81
f(z0) → n__f(z0) 0.00/0.81
p(s(0)) → 0 0.00/0.81
activate(n__f(z0)) → f(z0) 0.00/0.81
activate(z0) → z0
Tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0))) 0.00/0.81
ACTIVATE(n__f(z0)) → c4(F(z0))
S tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0))) 0.00/0.81
ACTIVATE(n__f(z0)) → c4(F(z0))
K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

F, ACTIVATE

Compound Symbols:

c1, c4

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

Removed 1 trailing tuple parts
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0))) 0.00/0.81
f(s(0)) → f(p(s(0))) 0.00/0.81
f(z0) → n__f(z0) 0.00/0.81
p(s(0)) → 0 0.00/0.81
activate(n__f(z0)) → f(z0) 0.00/0.81
activate(z0) → z0
Tuples:

ACTIVATE(n__f(z0)) → c4(F(z0)) 0.00/0.81
F(s(0)) → c1(F(p(s(0))))
S tuples:

ACTIVATE(n__f(z0)) → c4(F(z0)) 0.00/0.81
F(s(0)) → c1(F(p(s(0))))
K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

ACTIVATE, F

Compound Symbols:

c4, c1

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(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

ACTIVATE(n__f(z0)) → c4(F(z0))
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0))) 0.00/0.81
f(s(0)) → f(p(s(0))) 0.00/0.81
f(z0) → n__f(z0) 0.00/0.81
p(s(0)) → 0 0.00/0.81
activate(n__f(z0)) → f(z0) 0.00/0.81
activate(z0) → z0
Tuples:

F(s(0)) → c1(F(p(s(0))))
S tuples:

F(s(0)) → c1(F(p(s(0))))
K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

F

Compound Symbols:

c1

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0.00/0.81

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F(s(0)) → c1(F(p(s(0))))
We considered the (Usable) Rules:

p(s(0)) → 0
And the Tuples:

F(s(0)) → c1(F(p(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.81

POL(0) = [3]    0.00/0.81
POL(F(x1)) = [4]x1    0.00/0.81
POL(c1(x1)) = x1    0.00/0.81
POL(p(x1)) = [4]    0.00/0.81
POL(s(x1)) = [5] + x1   
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0))) 0.00/0.81
f(s(0)) → f(p(s(0))) 0.00/0.81
f(z0) → n__f(z0) 0.00/0.81
p(s(0)) → 0 0.00/0.81
activate(n__f(z0)) → f(z0) 0.00/0.81
activate(z0) → z0
Tuples:

F(s(0)) → c1(F(p(s(0))))
S tuples:none
K tuples:

F(s(0)) → c1(F(p(s(0))))
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

F

Compound Symbols:

c1

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

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

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0.00/0.83 EOF