YES(O(1), O(n^1)) 0.00/0.71 YES(O(1), O(n^1)) 0.00/0.72 0.00/0.72 0.00/0.72 0.00/0.72 0.00/0.72 0.00/0.72 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.72 0.00/0.72 0.00/0.72
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0.00/0.72
0.00/0.72

(0) Obligation:

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

g(s(x)) → f(x) 0.00/0.72
f(0) → s(0) 0.00/0.72
f(s(x)) → s(s(g(x))) 0.00/0.72
g(0) → 0

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

Converted CpxTRS to CDT
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0.00/0.72

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(s(z0)) → f(z0) 0.00/0.72
g(0) → 0 0.00/0.72
f(0) → s(0) 0.00/0.72
f(s(z0)) → s(s(g(z0)))
Tuples:

G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
S tuples:

G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
K tuples:none
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c, c3

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0.00/0.72

(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.

G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
We considered the (Usable) Rules:none
And the Tuples:

G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.72

POL(F(x1)) = [2] + [2]x1    0.00/0.72
POL(G(x1)) = [4] + [2]x1    0.00/0.72
POL(c(x1)) = x1    0.00/0.72
POL(c3(x1)) = x1    0.00/0.72
POL(s(x1)) = [4] + x1   
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0.00/0.72

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(s(z0)) → f(z0) 0.00/0.72
g(0) → 0 0.00/0.72
f(0) → s(0) 0.00/0.72
f(s(z0)) → s(s(g(z0)))
Tuples:

G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
S tuples:none
K tuples:

G(s(z0)) → c(F(z0)) 0.00/0.72
F(s(z0)) → c3(G(z0))
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c, c3

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0.00/0.72

(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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0.00/0.75 EOF