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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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))
0.00/0.87 
↳4 CdtProblem
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↳5 CdtNarrowingProof (BOTH BOUNDS(ID, ID))
0.00/0.87 
↳6 CdtProblem
0.00/0.87 
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳8 CdtProblem
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↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳10 BOUNDS(O(1), O(1))
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0.00/0.87 0.00/0.87
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0.00/0.87

(0) Obligation:

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

-(0, y) → 0 0.00/0.87
-(x, 0) → x 0.00/0.87
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0) 0.00/0.87
p(0) → 0 0.00/0.87
p(s(x)) → x

Rewrite Strategy: INNERMOST
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(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
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0.00/0.87

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(0, z0) → 0 0.00/0.87
-(z0, 0) → z0 0.00/0.87
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0) 0.00/0.87
p(0) → 0 0.00/0.87
p(s(z0)) → z0
Tuples:

-'(z0, s(z1)) → c2(-'(z0, p(s(z1))), P(s(z1)))
S tuples:

-'(z0, s(z1)) → c2(-'(z0, p(s(z1))), P(s(z1)))
K tuples:none
Defined Rule Symbols:

-, p

Defined Pair Symbols:

-'

Compound Symbols:

c2

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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:

-(0, z0) → 0 0.00/0.87
-(z0, 0) → z0 0.00/0.87
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0) 0.00/0.87
p(0) → 0 0.00/0.87
p(s(z0)) → z0
Tuples:

-'(z0, s(z1)) → c2(-'(z0, p(s(z1))))
S tuples:

-'(z0, s(z1)) → c2(-'(z0, p(s(z1))))
K tuples:none
Defined Rule Symbols:

-, p

Defined Pair Symbols:

-'

Compound Symbols:

c2

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0.00/0.87

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

Use narrowing to replace -'(z0, s(z1)) → c2(-'(z0, p(s(z1)))) by

-'(x0, s(z0)) → c2(-'(x0, z0))
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0.00/0.87

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(0, z0) → 0 0.00/0.87
-(z0, 0) → z0 0.00/0.87
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0) 0.00/0.87
p(0) → 0 0.00/0.87
p(s(z0)) → z0
Tuples:

-'(x0, s(z0)) → c2(-'(x0, z0))
S tuples:

-'(x0, s(z0)) → c2(-'(x0, z0))
K tuples:none
Defined Rule Symbols:

-, p

Defined Pair Symbols:

-'

Compound Symbols:

c2

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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.

-'(x0, s(z0)) → c2(-'(x0, z0))
We considered the (Usable) Rules:none
And the Tuples:

-'(x0, s(z0)) → c2(-'(x0, z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.87

POL(-'(x1, x2)) = [3]x2    0.00/0.87
POL(c2(x1)) = x1    0.00/0.87
POL(s(x1)) = [1] + x1   
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

-(0, z0) → 0 0.00/0.87
-(z0, 0) → z0 0.00/0.87
-(z0, s(z1)) → if(greater(z0, s(z1)), s(-(z0, p(s(z1)))), 0) 0.00/0.87
p(0) → 0 0.00/0.87
p(s(z0)) → z0
Tuples:

-'(x0, s(z0)) → c2(-'(x0, z0))
S tuples:none
K tuples:

-'(x0, s(z0)) → c2(-'(x0, z0))
Defined Rule Symbols:

-, p

Defined Pair Symbols:

-'

Compound Symbols:

c2

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

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

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0.00/0.89 EOF