YES(O(1), O(n^1)) 0.00/0.71 YES(O(1), O(n^1)) 0.00/0.73 0.00/0.73 0.00/0.73 0.00/0.73 0.00/0.73 0.00/0.73 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.73 0.00/0.73 0.00/0.73
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(0) Obligation:

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

f(c(X, s(Y))) → f(c(s(X), Y)) 0.00/0.73
g(c(s(X), Y)) → f(c(X, s(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(c(z0, s(z1))) → f(c(s(z0), z1)) 0.00/0.73
g(c(s(z0), z1)) → f(c(z0, s(z1)))
Tuples:

F(c(z0, s(z1))) → c1(F(c(s(z0), z1))) 0.00/0.73
G(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
S tuples:

F(c(z0, s(z1))) → c1(F(c(s(z0), z1))) 0.00/0.73
G(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

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(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

G(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(z0, s(z1))) → f(c(s(z0), z1)) 0.00/0.73
g(c(s(z0), z1)) → f(c(z0, s(z1)))
Tuples:

F(c(z0, s(z1))) → c1(F(c(s(z0), z1)))
S tuples:

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

f, g

Defined Pair Symbols:

F

Compound Symbols:

c1

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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.

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

F(c(z0, s(z1))) → c1(F(c(s(z0), z1)))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.73

POL(F(x1)) = x1    0.00/0.73
POL(c(x1, x2)) = x2    0.00/0.73
POL(c1(x1)) = x1    0.00/0.73
POL(s(x1)) = [2] + x1   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(c(z0, s(z1))) → f(c(s(z0), z1)) 0.00/0.73
g(c(s(z0), z1)) → f(c(z0, s(z1)))
Tuples:

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

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

f, g

Defined Pair Symbols:

F

Compound Symbols:

c1

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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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0.00/0.77 EOF