YES(O(1), O(n^2)) 0.00/0.83 YES(O(1), O(n^2)) 0.00/0.85 0.00/0.85 0.00/0.85 0.00/0.85 0.00/0.85 0.00/0.85 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.85 0.00/0.85 0.00/0.85
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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 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
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↳4 CdtProblem
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↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
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↳6 CdtProblem
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↳7 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳8 BOUNDS(O(1), O(1))
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0.00/0.85 0.00/0.85
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(0) Obligation:

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

*(x, *(y, z)) → *(otimes(x, y), z) 0.00/0.85
*(1, y) → y 0.00/0.85
*(+(x, y), z) → oplus(*(x, z), *(y, z)) 0.00/0.85
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))

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:

*(z0, *(z1, z2)) → *(otimes(z0, z1), z2) 0.00/0.85
*(1, z0) → z0 0.00/0.85
*(+(z0, z1), z2) → oplus(*(z0, z2), *(z1, z2)) 0.00/0.85
*(z0, oplus(z1, z2)) → oplus(*(z0, z1), *(z0, z2))
Tuples:

*'(z0, *(z1, z2)) → c(*'(otimes(z0, z1), z2)) 0.00/0.85
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2)) 0.00/0.85
*'(z0, oplus(z1, z2)) → c3(*'(z0, z1), *'(z0, z2))
S tuples:

*'(z0, *(z1, z2)) → c(*'(otimes(z0, z1), z2)) 0.00/0.85
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2)) 0.00/0.85
*'(z0, oplus(z1, z2)) → c3(*'(z0, z1), *'(z0, z2))
K tuples:none
Defined Rule Symbols:

*

Defined Pair Symbols:

*'

Compound Symbols:

c, c2, c3

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(3) 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.

*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2)) 0.00/0.85
*'(z0, oplus(z1, z2)) → c3(*'(z0, z1), *'(z0, z2))
We considered the (Usable) Rules:none
And the Tuples:

*'(z0, *(z1, z2)) → c(*'(otimes(z0, z1), z2)) 0.00/0.85
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2)) 0.00/0.85
*'(z0, oplus(z1, z2)) → c3(*'(z0, z1), *'(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.85

POL(*(x1, x2)) = [3]x1 + [2]x2 + [3]x1·x2    0.00/0.85
POL(*'(x1, x2)) = x1 + [2]x2 + x1·x2    0.00/0.85
POL(+(x1, x2)) = [2] + x1 + x2    0.00/0.85
POL(c(x1)) = x1    0.00/0.85
POL(c2(x1, x2)) = x1 + x2    0.00/0.85
POL(c3(x1, x2)) = x1 + x2    0.00/0.85
POL(oplus(x1, x2)) = [1] + x1 + x2    0.00/0.85
POL(otimes(x1, x2)) = x1 + x2   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

*(z0, *(z1, z2)) → *(otimes(z0, z1), z2) 0.00/0.85
*(1, z0) → z0 0.00/0.85
*(+(z0, z1), z2) → oplus(*(z0, z2), *(z1, z2)) 0.00/0.85
*(z0, oplus(z1, z2)) → oplus(*(z0, z1), *(z0, z2))
Tuples:

*'(z0, *(z1, z2)) → c(*'(otimes(z0, z1), z2)) 0.00/0.85
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2)) 0.00/0.85
*'(z0, oplus(z1, z2)) → c3(*'(z0, z1), *'(z0, z2))
S tuples:

*'(z0, *(z1, z2)) → c(*'(otimes(z0, z1), z2))
K tuples:

*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2)) 0.00/0.85
*'(z0, oplus(z1, z2)) → c3(*'(z0, z1), *'(z0, z2))
Defined Rule Symbols:

*

Defined Pair Symbols:

*'

Compound Symbols:

c, c2, c3

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(5) 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.

*'(z0, *(z1, z2)) → c(*'(otimes(z0, z1), z2))
We considered the (Usable) Rules:none
And the Tuples:

*'(z0, *(z1, z2)) → c(*'(otimes(z0, z1), z2)) 0.00/0.85
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2)) 0.00/0.85
*'(z0, oplus(z1, z2)) → c3(*'(z0, z1), *'(z0, z2))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.85

POL(*(x1, x2)) = [3] + [3]x1 + [2]x2 + [3]x1·x2    0.00/0.85
POL(*'(x1, x2)) = [3] + x1 + [3]x2 + x1·x2    0.00/0.85
POL(+(x1, x2)) = [3] + x1 + x2    0.00/0.85
POL(c(x1)) = x1    0.00/0.85
POL(c2(x1, x2)) = x1 + x2    0.00/0.85
POL(c3(x1, x2)) = x1 + x2    0.00/0.85
POL(oplus(x1, x2)) = [3] + x1 + x2    0.00/0.85
POL(otimes(x1, x2)) = [1] + x1 + x2   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

*(z0, *(z1, z2)) → *(otimes(z0, z1), z2) 0.00/0.85
*(1, z0) → z0 0.00/0.85
*(+(z0, z1), z2) → oplus(*(z0, z2), *(z1, z2)) 0.00/0.85
*(z0, oplus(z1, z2)) → oplus(*(z0, z1), *(z0, z2))
Tuples:

*'(z0, *(z1, z2)) → c(*'(otimes(z0, z1), z2)) 0.00/0.85
*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2)) 0.00/0.85
*'(z0, oplus(z1, z2)) → c3(*'(z0, z1), *'(z0, z2))
S tuples:none
K tuples:

*'(+(z0, z1), z2) → c2(*'(z0, z2), *'(z1, z2)) 0.00/0.85
*'(z0, oplus(z1, z2)) → c3(*'(z0, z1), *'(z0, z2)) 0.00/0.85
*'(z0, *(z1, z2)) → c(*'(otimes(z0, z1), z2))
Defined Rule Symbols:

*

Defined Pair Symbols:

*'

Compound Symbols:

c, c2, c3

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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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0.00/0.89 EOF