YES(O(1), O(n^1)) 0.00/0.70 YES(O(1), O(n^1)) 0.00/0.71 0.00/0.71 0.00/0.71 0.00/0.71 0.00/0.71 0.00/0.71 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.71 0.00/0.71 0.00/0.71
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The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(O(1), O(n^1)).

0 CpxRelTRS
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↳1 CpxRelTrsToCDT (UPPER BOUND (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 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳6 CdtProblem
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↳7 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳8 BOUNDS(O(1), O(1))
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(0) Obligation:

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

div2(S(S(x))) → +(S(0), div2(x)) 0.00/0.71
div2(S(0)) → 0 0.00/0.71
div2(0) → 0

The (relative) TRS S consists of the following rules:

+(x, S(0)) → S(x) 0.00/0.71
+(S(0), y) → S(y)

Rewrite Strategy: INNERMOST
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(1) CpxRelTrsToCDT (UPPER BOUND (ID) transformation)

Relative innermost TRS to CDT Problem.
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(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, S(0)) → S(z0) 0.00/0.71
+(S(0), z0) → S(z0) 0.00/0.71
div2(S(S(z0))) → +(S(0), div2(z0)) 0.00/0.71
div2(S(0)) → 0 0.00/0.71
div2(0) → 0
Tuples:

DIV2(S(S(z0))) → c2(+'(S(0), div2(z0)), DIV2(z0))
S tuples:

DIV2(S(S(z0))) → c2(+'(S(0), div2(z0)), DIV2(z0))
K tuples:none
Defined Rule Symbols:

div2, +

Defined Pair Symbols:

DIV2

Compound Symbols:

c2

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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:

+(z0, S(0)) → S(z0) 0.00/0.71
+(S(0), z0) → S(z0) 0.00/0.71
div2(S(S(z0))) → +(S(0), div2(z0)) 0.00/0.71
div2(S(0)) → 0 0.00/0.71
div2(0) → 0
Tuples:

DIV2(S(S(z0))) → c2(DIV2(z0))
S tuples:

DIV2(S(S(z0))) → c2(DIV2(z0))
K tuples:none
Defined Rule Symbols:

div2, +

Defined Pair Symbols:

DIV2

Compound Symbols:

c2

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(5) 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.

DIV2(S(S(z0))) → c2(DIV2(z0))
We considered the (Usable) Rules:none
And the Tuples:

DIV2(S(S(z0))) → c2(DIV2(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.71

POL(DIV2(x1)) = [4]x1    0.00/0.71
POL(S(x1)) = [4] + x1    0.00/0.71
POL(c2(x1)) = x1   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, S(0)) → S(z0) 0.00/0.71
+(S(0), z0) → S(z0) 0.00/0.71
div2(S(S(z0))) → +(S(0), div2(z0)) 0.00/0.71
div2(S(0)) → 0 0.00/0.71
div2(0) → 0
Tuples:

DIV2(S(S(z0))) → c2(DIV2(z0))
S tuples:none
K tuples:

DIV2(S(S(z0))) → c2(DIV2(z0))
Defined Rule Symbols:

div2, +

Defined Pair Symbols:

DIV2

Compound Symbols:

c2

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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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0.00/0.73 EOF