YES(O(1), O(n^1)) 0.00/0.86 YES(O(1), O(n^1)) 0.00/0.89 0.00/0.89 0.00/0.89 0.00/0.89 0.00/0.89 0.00/0.89 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.89 0.00/0.89 0.00/0.89
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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 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID))
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↳4 CdtProblem
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↳5 CdtLeafRemovalProof (ComplexityIfPolyImplication)
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↳6 CdtProblem
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↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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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:

perfectp(0) → false 0.00/0.89
perfectp(s(x)) → f(x, s(0), s(x), s(x)) 0.00/0.89
f(0, y, 0, u) → true 0.00/0.89
f(0, y, s(z), u) → false 0.00/0.89
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u) 0.00/0.89
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

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:

perfectp(0) → false 0.00/0.89
perfectp(s(z0)) → f(z0, s(0), s(z0), s(z0)) 0.00/0.89
f(0, z0, 0, z1) → true 0.00/0.89
f(0, z0, s(z1), z2) → false 0.00/0.89
f(s(z0), 0, z1, z2) → f(z0, z2, minus(z1, s(z0)), z2) 0.00/0.89
f(s(z0), s(z1), z2, z3) → if(le(z0, z1), f(s(z0), minus(z1, z0), z2, z3), f(z0, z3, z2, z3))
Tuples:

PERFECTP(s(z0)) → c1(F(z0, s(0), s(z0), s(z0))) 0.00/0.89
F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2)) 0.00/0.89
F(s(z0), s(z1), z2, z3) → c5(F(s(z0), minus(z1, z0), z2, z3), F(z0, z3, z2, z3))
S tuples:

PERFECTP(s(z0)) → c1(F(z0, s(0), s(z0), s(z0))) 0.00/0.89
F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2)) 0.00/0.89
F(s(z0), s(z1), z2, z3) → c5(F(s(z0), minus(z1, z0), z2, z3), F(z0, z3, z2, z3))
K tuples:none
Defined Rule Symbols:

perfectp, f

Defined Pair Symbols:

PERFECTP, F

Compound Symbols:

c1, c4, c5

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

Removed 1 trailing tuple parts
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

perfectp(0) → false 0.00/0.89
perfectp(s(z0)) → f(z0, s(0), s(z0), s(z0)) 0.00/0.89
f(0, z0, 0, z1) → true 0.00/0.89
f(0, z0, s(z1), z2) → false 0.00/0.89
f(s(z0), 0, z1, z2) → f(z0, z2, minus(z1, s(z0)), z2) 0.00/0.89
f(s(z0), s(z1), z2, z3) → if(le(z0, z1), f(s(z0), minus(z1, z0), z2, z3), f(z0, z3, z2, z3))
Tuples:

PERFECTP(s(z0)) → c1(F(z0, s(0), s(z0), s(z0))) 0.00/0.89
F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2)) 0.00/0.89
F(s(z0), s(z1), z2, z3) → c5(F(z0, z3, z2, z3))
S tuples:

PERFECTP(s(z0)) → c1(F(z0, s(0), s(z0), s(z0))) 0.00/0.89
F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2)) 0.00/0.89
F(s(z0), s(z1), z2, z3) → c5(F(z0, z3, z2, z3))
K tuples:none
Defined Rule Symbols:

perfectp, f

Defined Pair Symbols:

PERFECTP, F

Compound Symbols:

c1, c4, c5

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(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

PERFECTP(s(z0)) → c1(F(z0, s(0), s(z0), s(z0)))
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

perfectp(0) → false 0.00/0.89
perfectp(s(z0)) → f(z0, s(0), s(z0), s(z0)) 0.00/0.89
f(0, z0, 0, z1) → true 0.00/0.89
f(0, z0, s(z1), z2) → false 0.00/0.89
f(s(z0), 0, z1, z2) → f(z0, z2, minus(z1, s(z0)), z2) 0.00/0.89
f(s(z0), s(z1), z2, z3) → if(le(z0, z1), f(s(z0), minus(z1, z0), z2, z3), f(z0, z3, z2, z3))
Tuples:

F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2)) 0.00/0.89
F(s(z0), s(z1), z2, z3) → c5(F(z0, z3, z2, z3))
S tuples:

F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2)) 0.00/0.89
F(s(z0), s(z1), z2, z3) → c5(F(z0, z3, z2, z3))
K tuples:none
Defined Rule Symbols:

perfectp, f

Defined Pair Symbols:

F

Compound Symbols:

c4, c5

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(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.

F(s(z0), s(z1), z2, z3) → c5(F(z0, z3, z2, z3))
We considered the (Usable) Rules:none
And the Tuples:

F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2)) 0.00/0.89
F(s(z0), s(z1), z2, z3) → c5(F(z0, z3, z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.89

POL(0) = 0    0.00/0.89
POL(F(x1, x2, x3, x4)) = [4]x1 + [4]x3    0.00/0.89
POL(c4(x1)) = x1    0.00/0.89
POL(c5(x1)) = x1    0.00/0.89
POL(minus(x1, x2)) = [4] + x1    0.00/0.89
POL(s(x1)) = [4] + x1   
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

perfectp(0) → false 0.00/0.89
perfectp(s(z0)) → f(z0, s(0), s(z0), s(z0)) 0.00/0.89
f(0, z0, 0, z1) → true 0.00/0.89
f(0, z0, s(z1), z2) → false 0.00/0.89
f(s(z0), 0, z1, z2) → f(z0, z2, minus(z1, s(z0)), z2) 0.00/0.89
f(s(z0), s(z1), z2, z3) → if(le(z0, z1), f(s(z0), minus(z1, z0), z2, z3), f(z0, z3, z2, z3))
Tuples:

F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2)) 0.00/0.89
F(s(z0), s(z1), z2, z3) → c5(F(z0, z3, z2, z3))
S tuples:

F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2))
K tuples:

F(s(z0), s(z1), z2, z3) → c5(F(z0, z3, z2, z3))
Defined Rule Symbols:

perfectp, f

Defined Pair Symbols:

F

Compound Symbols:

c4, c5

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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), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2))
We considered the (Usable) Rules:none
And the Tuples:

F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2)) 0.00/0.89
F(s(z0), s(z1), z2, z3) → c5(F(z0, z3, z2, z3))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.89

POL(0) = 0    0.00/0.89
POL(F(x1, x2, x3, x4)) = [4]x1    0.00/0.89
POL(c4(x1)) = x1    0.00/0.89
POL(c5(x1)) = x1    0.00/0.89
POL(minus(x1, x2)) = [1] + x1 + x2    0.00/0.89
POL(s(x1)) = [4] + x1   
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(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

perfectp(0) → false 0.00/0.89
perfectp(s(z0)) → f(z0, s(0), s(z0), s(z0)) 0.00/0.89
f(0, z0, 0, z1) → true 0.00/0.89
f(0, z0, s(z1), z2) → false 0.00/0.89
f(s(z0), 0, z1, z2) → f(z0, z2, minus(z1, s(z0)), z2) 0.00/0.89
f(s(z0), s(z1), z2, z3) → if(le(z0, z1), f(s(z0), minus(z1, z0), z2, z3), f(z0, z3, z2, z3))
Tuples:

F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2)) 0.00/0.89
F(s(z0), s(z1), z2, z3) → c5(F(z0, z3, z2, z3))
S tuples:none
K tuples:

F(s(z0), s(z1), z2, z3) → c5(F(z0, z3, z2, z3)) 0.00/0.89
F(s(z0), 0, z1, z2) → c4(F(z0, z2, minus(z1, s(z0)), z2))
Defined Rule Symbols:

perfectp, f

Defined Pair Symbols:

F

Compound Symbols:

c4, c5

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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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0.00/0.94 EOF