YES(O(1), O(n^1)) 3.04/1.26 YES(O(1), O(n^1)) 3.04/1.28 3.04/1.28 3.04/1.28 3.04/1.28 3.04/1.28 3.04/1.28 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 3.04/1.28 3.04/1.28 3.04/1.28
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The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

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

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

f(x, c(y)) → f(x, s(f(y, y))) 3.04/1.28
f(s(x), y) → f(x, s(c(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, c(z1)) → f(z0, s(f(z1, z1))) 3.04/1.28
f(s(z0), z1) → f(z0, s(c(z1)))
Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1)) 3.04/1.28
F(s(z0), z1) → c2(F(z0, s(c(z1))))
S tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1)) 3.04/1.28
F(s(z0), z1) → c2(F(z0, s(c(z1))))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2

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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, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1))
We considered the (Usable) Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1))) 3.04/1.28
f(s(z0), z1) → f(z0, s(c(z1)))
And the Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1)) 3.04/1.28
F(s(z0), z1) → c2(F(z0, s(c(z1))))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.04/1.28

POL(F(x1, x2)) = [2]x2    3.04/1.28
POL(c(x1)) = [1] + x1    3.04/1.28
POL(c1(x1, x2)) = x1 + x2    3.04/1.28
POL(c2(x1)) = x1    3.04/1.28
POL(f(x1, x2)) = [5] + x1    3.04/1.28
POL(s(x1)) = 0   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1))) 3.04/1.28
f(s(z0), z1) → f(z0, s(c(z1)))
Tuples:

F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1)) 3.04/1.28
F(s(z0), z1) → c2(F(z0, s(c(z1))))
S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
K tuples:

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

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c2

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

Use instantiation to replace F(z0, c(z1)) → c1(F(z0, s(f(z1, z1))), F(z1, z1)) by

F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1))) 3.04/1.28
f(s(z0), z1) → f(z0, s(c(z1)))
Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1)))) 3.04/1.28
F(c(z1), c(z1)) → c1(F(c(z1), s(f(z1, z1))), F(z1, z1))
S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
K tuples:

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

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

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

Removed 1 trailing tuple parts
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1))) 3.04/1.28
f(s(z0), z1) → f(z0, s(c(z1)))
Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1)))) 3.04/1.28
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1))))
K tuples:

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

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

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(9) 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(s(z0), z1) → c2(F(z0, s(c(z1))))
We considered the (Usable) Rules:none
And the Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1)))) 3.04/1.28
F(c(z1), c(z1)) → c1(F(z1, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 3.04/1.28

POL(F(x1, x2)) = [4]x1    3.04/1.28
POL(c(x1)) = [1] + x1    3.04/1.28
POL(c1(x1)) = x1    3.04/1.28
POL(c2(x1)) = x1    3.04/1.28
POL(s(x1)) = [4] + x1   
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(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0, c(z1)) → f(z0, s(f(z1, z1))) 3.04/1.28
f(s(z0), z1) → f(z0, s(c(z1)))
Tuples:

F(s(z0), z1) → c2(F(z0, s(c(z1)))) 3.04/1.28
F(c(z1), c(z1)) → c1(F(z1, z1))
S tuples:none
K tuples:

F(c(z1), c(z1)) → c1(F(z1, z1)) 3.04/1.28
F(s(z0), z1) → c2(F(z0, s(c(z1))))
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c1

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

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

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3.33/1.36 EOF