YES(O(1), O(n^1)) 0.00/0.75 YES(O(1), O(n^1)) 0.00/0.76 0.00/0.76 0.00/0.76 0.00/0.76 0.00/0.76 0.00/0.76 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.76 0.00/0.76 0.00/0.76
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(0) Obligation:

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

g(f(x), y) → f(h(x, y)) 0.00/0.76
h(x, y) → g(x, f(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:

g(f(z0), z1) → f(h(z0, z1)) 0.00/0.76
h(z0, z1) → g(z0, f(z1))
Tuples:

G(f(z0), z1) → c(H(z0, z1)) 0.00/0.76
H(z0, z1) → c1(G(z0, f(z1)))
S tuples:

G(f(z0), z1) → c(H(z0, z1)) 0.00/0.76
H(z0, z1) → c1(G(z0, f(z1)))
K tuples:none
Defined Rule Symbols:

g, h

Defined Pair Symbols:

G, H

Compound Symbols:

c, 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.

G(f(z0), z1) → c(H(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

G(f(z0), z1) → c(H(z0, z1)) 0.00/0.76
H(z0, z1) → c1(G(z0, f(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.76

POL(G(x1, x2)) = x1    0.00/0.76
POL(H(x1, x2)) = x1    0.00/0.76
POL(c(x1)) = x1    0.00/0.76
POL(c1(x1)) = x1    0.00/0.76
POL(f(x1)) = [1] + x1   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(f(z0), z1) → f(h(z0, z1)) 0.00/0.76
h(z0, z1) → g(z0, f(z1))
Tuples:

G(f(z0), z1) → c(H(z0, z1)) 0.00/0.76
H(z0, z1) → c1(G(z0, f(z1)))
S tuples:

H(z0, z1) → c1(G(z0, f(z1)))
K tuples:

G(f(z0), z1) → c(H(z0, z1))
Defined Rule Symbols:

g, h

Defined Pair Symbols:

G, H

Compound Symbols:

c, c1

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(5) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

H(z0, z1) → c1(G(z0, f(z1))) 0.00/0.76
G(f(z0), z1) → c(H(z0, z1))
Now S is empty
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(6) BOUNDS(O(1), O(1))

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0.00/0.80 EOF