YES(O(1), O(n^1)) 0.00/0.86 YES(O(1), O(n^1)) 0.00/0.87 0.00/0.87 0.00/0.87 0.00/0.87 0.00/0.87 0.00/0.87 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.87 0.00/0.87 0.00/0.87
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0.00/0.87
0.00/0.87

(0) Obligation:

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

p(s(x)) → x 0.00/0.87
fac(0) → s(0) 0.00/0.87
fac(s(x)) → times(s(x), fac(p(s(x))))

Rewrite Strategy: INNERMOST
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0.00/0.87

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
0.00/0.87
0.00/0.87

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0 0.00/0.87
fac(0) → s(0) 0.00/0.87
fac(s(z0)) → times(s(z0), fac(p(s(z0))))
Tuples:

FAC(s(z0)) → c2(FAC(p(s(z0))), P(s(z0)))
S tuples:

FAC(s(z0)) → c2(FAC(p(s(z0))), P(s(z0)))
K tuples:none
Defined Rule Symbols:

p, fac

Defined Pair Symbols:

FAC

Compound Symbols:

c2

0.00/0.87
0.00/0.87

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

Removed 1 trailing tuple parts
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0.00/0.87

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0 0.00/0.87
fac(0) → s(0) 0.00/0.87
fac(s(z0)) → times(s(z0), fac(p(s(z0))))
Tuples:

FAC(s(z0)) → c2(FAC(p(s(z0))))
S tuples:

FAC(s(z0)) → c2(FAC(p(s(z0))))
K tuples:none
Defined Rule Symbols:

p, fac

Defined Pair Symbols:

FAC

Compound Symbols:

c2

0.00/0.87
0.00/0.87

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

Use narrowing to replace FAC(s(z0)) → c2(FAC(p(s(z0)))) by

FAC(s(z0)) → c2(FAC(z0))
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0 0.00/0.87
fac(0) → s(0) 0.00/0.87
fac(s(z0)) → times(s(z0), fac(p(s(z0))))
Tuples:

FAC(s(z0)) → c2(FAC(z0))
S tuples:

FAC(s(z0)) → c2(FAC(z0))
K tuples:none
Defined Rule Symbols:

p, fac

Defined Pair Symbols:

FAC

Compound Symbols:

c2

0.00/0.87
0.00/0.87

(7) 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.

FAC(s(z0)) → c2(FAC(z0))
We considered the (Usable) Rules:none
And the Tuples:

FAC(s(z0)) → c2(FAC(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.87

POL(FAC(x1)) = [3]x1    0.00/0.87
POL(c2(x1)) = x1    0.00/0.87
POL(s(x1)) = [1] + x1   
0.00/0.87
0.00/0.87

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0 0.00/0.87
fac(0) → s(0) 0.00/0.87
fac(s(z0)) → times(s(z0), fac(p(s(z0))))
Tuples:

FAC(s(z0)) → c2(FAC(z0))
S tuples:none
K tuples:

FAC(s(z0)) → c2(FAC(z0))
Defined Rule Symbols:

p, fac

Defined Pair Symbols:

FAC

Compound Symbols:

c2

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0.00/0.87

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

The set S is empty
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0.00/0.87

(10) BOUNDS(O(1), O(1))

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0.00/0.90 EOF