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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(0) Obligation:

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

f(x, a) → x 0.00/0.71
f(x, g(y)) → f(g(x), y)

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:

f(z0, a) → z0 0.00/0.71
f(z0, g(z1)) → f(g(z0), z1)
Tuples:

F(z0, g(z1)) → c1(F(g(z0), z1))
S tuples:

F(z0, g(z1)) → c1(F(g(z0), z1))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

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(3) 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(z0, g(z1)) → c1(F(g(z0), z1))
We considered the (Usable) Rules:none
And the Tuples:

F(z0, g(z1)) → c1(F(g(z0), z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.71

POL(F(x1, x2)) = x2    0.00/0.71
POL(c1(x1)) = x1    0.00/0.71
POL(g(x1)) = [1] + x1   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, a) → z0 0.00/0.71
f(z0, g(z1)) → f(g(z0), z1)
Tuples:

F(z0, g(z1)) → c1(F(g(z0), z1))
S tuples:none
K tuples:

F(z0, g(z1)) → c1(F(g(z0), z1))
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

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

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

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0.00/0.73 EOF