YES(O(1), O(n^2)) 0.00/0.81 YES(O(1), O(n^2)) 0.00/0.83 0.00/0.83 0.00/0.83 0.00/0.83 0.00/0.83 0.00/0.83 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.83 0.00/0.83 0.00/0.83
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

0 CpxTRS
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↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
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↳2 CdtProblem
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↳3 CdtUnreachableProof (⇔)
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↳4 CdtProblem
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↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
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↳6 CdtProblem
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↳7 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
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↳8 CdtProblem
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↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
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↳10 CdtProblem
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↳11 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳12 CdtProblem
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↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳14 BOUNDS(O(1), O(1))
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(0) Obligation:

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

f(a) → g(h(a)) 0.00/0.83
h(g(x)) → g(h(f(x))) 0.00/0.83
k(x, h(x), a) → h(x) 0.00/0.83
k(f(x), y, x) → f(x)

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(a) → g(h(a)) 0.00/0.83
h(g(z0)) → g(h(f(z0))) 0.00/0.83
k(z0, h(z0), a) → h(z0) 0.00/0.83
k(f(z0), z1, z0) → f(z0)
Tuples:

F(a) → c(H(a)) 0.00/0.83
H(g(z0)) → c1(H(f(z0)), F(z0)) 0.00/0.83
K(z0, h(z0), a) → c2(H(z0)) 0.00/0.83
K(f(z0), z1, z0) → c3(F(z0))
S tuples:

F(a) → c(H(a)) 0.00/0.83
H(g(z0)) → c1(H(f(z0)), F(z0)) 0.00/0.83
K(z0, h(z0), a) → c2(H(z0)) 0.00/0.83
K(f(z0), z1, z0) → c3(F(z0))
K tuples:none
Defined Rule Symbols:

f, h, k

Defined Pair Symbols:

F, H, K

Compound Symbols:

c, c1, c2, c3

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(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

K(z0, h(z0), a) → c2(H(z0)) 0.00/0.83
K(f(z0), z1, z0) → c3(F(z0))
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → g(h(a)) 0.00/0.83
h(g(z0)) → g(h(f(z0))) 0.00/0.83
k(z0, h(z0), a) → h(z0) 0.00/0.83
k(f(z0), z1, z0) → f(z0)
Tuples:

F(a) → c(H(a)) 0.00/0.83
H(g(z0)) → c1(H(f(z0)), F(z0))
S tuples:

F(a) → c(H(a)) 0.00/0.83
H(g(z0)) → c1(H(f(z0)), F(z0))
K tuples:none
Defined Rule Symbols:

f, h, k

Defined Pair Symbols:

F, H

Compound Symbols:

c, c1

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

Removed 1 trailing tuple parts
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → g(h(a)) 0.00/0.83
h(g(z0)) → g(h(f(z0))) 0.00/0.83
k(z0, h(z0), a) → h(z0) 0.00/0.83
k(f(z0), z1, z0) → f(z0)
Tuples:

H(g(z0)) → c1(H(f(z0)), F(z0)) 0.00/0.83
F(a) → c
S tuples:

H(g(z0)) → c1(H(f(z0)), F(z0)) 0.00/0.83
F(a) → c
K tuples:none
Defined Rule Symbols:

f, h, k

Defined Pair Symbols:

H, F

Compound Symbols:

c1, c

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(7) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing nodes:

F(a) → c
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → g(h(a)) 0.00/0.83
h(g(z0)) → g(h(f(z0))) 0.00/0.83
k(z0, h(z0), a) → h(z0) 0.00/0.83
k(f(z0), z1, z0) → f(z0)
Tuples:

H(g(z0)) → c1(H(f(z0)), F(z0)) 0.00/0.83
F(a) → c
S tuples:

H(g(z0)) → c1(H(f(z0)), F(z0)) 0.00/0.83
F(a) → c
K tuples:none
Defined Rule Symbols:

f, h, k

Defined Pair Symbols:

H, F

Compound Symbols:

c1, c

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(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

F(a) → c
We considered the (Usable) Rules:

f(a) → g(h(a))
And the Tuples:

H(g(z0)) → c1(H(f(z0)), F(z0)) 0.00/0.83
F(a) → c
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.83

POL(F(x1)) = x12    0.00/0.83
POL(H(x1)) = x1 + x12    0.00/0.83
POL(a) = [1]    0.00/0.83
POL(c) = 0    0.00/0.83
POL(c1(x1, x2)) = x1 + x2    0.00/0.83
POL(f(x1)) = 0    0.00/0.83
POL(g(x1)) = x1    0.00/0.83
POL(h(x1)) = 0   
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(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → g(h(a)) 0.00/0.83
h(g(z0)) → g(h(f(z0))) 0.00/0.83
k(z0, h(z0), a) → h(z0) 0.00/0.83
k(f(z0), z1, z0) → f(z0)
Tuples:

H(g(z0)) → c1(H(f(z0)), F(z0)) 0.00/0.83
F(a) → c
S tuples:

H(g(z0)) → c1(H(f(z0)), F(z0))
K tuples:

F(a) → c
Defined Rule Symbols:

f, h, k

Defined Pair Symbols:

H, F

Compound Symbols:

c1, c

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(11) 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.

H(g(z0)) → c1(H(f(z0)), F(z0))
We considered the (Usable) Rules:

f(a) → g(h(a))
And the Tuples:

H(g(z0)) → c1(H(f(z0)), F(z0)) 0.00/0.83
F(a) → c
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.83

POL(F(x1)) = [1]    0.00/0.83
POL(H(x1)) = [4]x1    0.00/0.83
POL(a) = [4]    0.00/0.83
POL(c) = 0    0.00/0.83
POL(c1(x1, x2)) = x1 + x2    0.00/0.83
POL(f(x1)) = x1    0.00/0.83
POL(g(x1)) = [4] + x1    0.00/0.83
POL(h(x1)) = 0   
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(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(a) → g(h(a)) 0.00/0.83
h(g(z0)) → g(h(f(z0))) 0.00/0.83
k(z0, h(z0), a) → h(z0) 0.00/0.83
k(f(z0), z1, z0) → f(z0)
Tuples:

H(g(z0)) → c1(H(f(z0)), F(z0)) 0.00/0.83
F(a) → c
S tuples:none
K tuples:

F(a) → c 0.00/0.83
H(g(z0)) → c1(H(f(z0)), F(z0))
Defined Rule Symbols:

f, h, k

Defined Pair Symbols:

H, F

Compound Symbols:

c1, c

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

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

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0.00/0.89 EOF