YES(O(1), O(n^1)) 0.00/0.78 YES(O(1), O(n^1)) 0.00/0.79 0.00/0.79 0.00/0.79 0.00/0.79 0.00/0.79 0.00/0.79 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.79 0.00/0.79 0.00/0.79
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxTRS
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↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
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↳2 CdtProblem
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↳3 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
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↳4 CdtProblem
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↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
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↳6 CdtProblem
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↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳8 CdtProblem
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↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳10 BOUNDS(O(1), O(1))
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0.00/0.79 0.00/0.79
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(0) Obligation:

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

from(X) → cons(X, n__from(s(X))) 0.00/0.79
after(0, XS) → XS 0.00/0.79
after(s(N), cons(X, XS)) → after(N, activate(XS)) 0.00/0.79
from(X) → n__from(X) 0.00/0.79
activate(n__from(X)) → from(X) 0.00/0.79
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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(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 0.00/0.79
from(z0) → n__from(z0) 0.00/0.79
after(0, z0) → z0 0.00/0.79
after(s(z0), cons(z1, z2)) → after(z0, activate(z2)) 0.00/0.79
activate(n__from(z0)) → from(z0) 0.00/0.79
activate(z0) → z0
Tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.79
ACTIVATE(n__from(z0)) → c4(FROM(z0))
S tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.79
ACTIVATE(n__from(z0)) → c4(FROM(z0))
K tuples:none
Defined Rule Symbols:

from, after, activate

Defined Pair Symbols:

AFTER, ACTIVATE

Compound Symbols:

c3, 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:

from(z0) → cons(z0, n__from(s(z0))) 0.00/0.79
from(z0) → n__from(z0) 0.00/0.79
after(0, z0) → z0 0.00/0.79
after(s(z0), cons(z1, z2)) → after(z0, activate(z2)) 0.00/0.79
activate(n__from(z0)) → from(z0) 0.00/0.79
activate(z0) → z0
Tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.79
ACTIVATE(n__from(z0)) → c4
S tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.79
ACTIVATE(n__from(z0)) → c4
K tuples:none
Defined Rule Symbols:

from, after, activate

Defined Pair Symbols:

AFTER, ACTIVATE

Compound Symbols:

c3, c4

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

Removed 1 trailing nodes:

ACTIVATE(n__from(z0)) → c4
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 0.00/0.79
from(z0) → n__from(z0) 0.00/0.79
after(0, z0) → z0 0.00/0.79
after(s(z0), cons(z1, z2)) → after(z0, activate(z2)) 0.00/0.79
activate(n__from(z0)) → from(z0) 0.00/0.79
activate(z0) → z0
Tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.79
ACTIVATE(n__from(z0)) → c4
S tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.79
ACTIVATE(n__from(z0)) → c4
K tuples:none
Defined Rule Symbols:

from, after, activate

Defined Pair Symbols:

AFTER, ACTIVATE

Compound Symbols:

c3, c4

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(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.

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.79
ACTIVATE(n__from(z0)) → c4
We considered the (Usable) Rules:

activate(n__from(z0)) → from(z0) 0.00/0.79
activate(z0) → z0 0.00/0.79
from(z0) → cons(z0, n__from(s(z0))) 0.00/0.79
from(z0) → n__from(z0)
And the Tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.79
ACTIVATE(n__from(z0)) → c4
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.79

POL(ACTIVATE(x1)) = [1]    0.00/0.79
POL(AFTER(x1, x2)) = [2]x1    0.00/0.79
POL(activate(x1)) = 0    0.00/0.79
POL(c3(x1, x2)) = x1 + x2    0.00/0.79
POL(c4) = 0    0.00/0.79
POL(cons(x1, x2)) = x1    0.00/0.79
POL(from(x1)) = [3] + [3]x1    0.00/0.79
POL(n__from(x1)) = x1    0.00/0.79
POL(s(x1)) = [1] + x1   
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0))) 0.00/0.79
from(z0) → n__from(z0) 0.00/0.79
after(0, z0) → z0 0.00/0.79
after(s(z0), cons(z1, z2)) → after(z0, activate(z2)) 0.00/0.79
activate(n__from(z0)) → from(z0) 0.00/0.79
activate(z0) → z0
Tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.79
ACTIVATE(n__from(z0)) → c4
S tuples:none
K tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2)) 0.00/0.79
ACTIVATE(n__from(z0)) → c4
Defined Rule Symbols:

from, after, activate

Defined Pair Symbols:

AFTER, ACTIVATE

Compound Symbols:

c3, c4

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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