YES(O(1), O(n^1)) 0.00/0.76 YES(O(1), O(n^1)) 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 0.00/0.77 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.77 0.00/0.77 0.00/0.77
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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 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳6 BOUNDS(O(1), O(1))
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0.00/0.77 0.00/0.77
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(0) Obligation:

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

D(t) → 1 0.00/0.77
D(constant) → 0 0.00/0.77
D(+(x, y)) → +(D(x), D(y)) 0.00/0.77
D(*(x, y)) → +(*(y, D(x)), *(x, D(y))) 0.00/0.77
D(-(x, y)) → -(D(x), D(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:

D(t) → 1 0.00/0.77
D(constant) → 0 0.00/0.77
D(+(z0, z1)) → +(D(z0), D(z1)) 0.00/0.77
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1))) 0.00/0.77
D(-(z0, z1)) → -(D(z0), D(z1))
Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1)) 0.00/0.77
D'(*(z0, z1)) → c3(D'(z0), D'(z1)) 0.00/0.77
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
S tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1)) 0.00/0.77
D'(*(z0, z1)) → c3(D'(z0), D'(z1)) 0.00/0.77
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
K tuples:none
Defined Rule Symbols:

D

Defined Pair Symbols:

D'

Compound Symbols:

c2, c3, c4

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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.

D'(+(z0, z1)) → c2(D'(z0), D'(z1)) 0.00/0.77
D'(*(z0, z1)) → c3(D'(z0), D'(z1)) 0.00/0.77
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
We considered the (Usable) Rules:none
And the Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1)) 0.00/0.77
D'(*(z0, z1)) → c3(D'(z0), D'(z1)) 0.00/0.77
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.77

POL(*(x1, x2)) = [3] + x1 + x2    0.00/0.77
POL(+(x1, x2)) = [1] + x1 + x2    0.00/0.77
POL(-(x1, x2)) = [3] + x1 + x2    0.00/0.77
POL(D'(x1)) = [2]x1    0.00/0.77
POL(c2(x1, x2)) = x1 + x2    0.00/0.77
POL(c3(x1, x2)) = x1 + x2    0.00/0.77
POL(c4(x1, x2)) = x1 + x2   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

D(t) → 1 0.00/0.77
D(constant) → 0 0.00/0.77
D(+(z0, z1)) → +(D(z0), D(z1)) 0.00/0.77
D(*(z0, z1)) → +(*(z1, D(z0)), *(z0, D(z1))) 0.00/0.77
D(-(z0, z1)) → -(D(z0), D(z1))
Tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1)) 0.00/0.77
D'(*(z0, z1)) → c3(D'(z0), D'(z1)) 0.00/0.77
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
S tuples:none
K tuples:

D'(+(z0, z1)) → c2(D'(z0), D'(z1)) 0.00/0.77
D'(*(z0, z1)) → c3(D'(z0), D'(z1)) 0.00/0.77
D'(-(z0, z1)) → c4(D'(z0), D'(z1))
Defined Rule Symbols:

D

Defined Pair Symbols:

D'

Compound Symbols:

c2, c3, c4

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(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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0.00/0.78 EOF