YES(O(1), O(n^1)) 0.00/0.70 YES(O(1), O(n^1)) 0.00/0.72 0.00/0.72 0.00/0.72 0.00/0.72 0.00/0.72 0.00/0.72 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.72 0.00/0.72 0.00/0.72
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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))
0.00/0.72 
↳4 CdtProblem
0.00/0.72 
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
0.00/0.72 
↳6 CdtProblem
0.00/0.72 
↳7 SIsEmptyProof (BOTH BOUNDS(ID, ID))
0.00/0.72 
↳8 BOUNDS(O(1), O(1))
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0.00/0.72 0.00/0.72
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0.00/0.72

(0) Obligation:

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

dbl(S(0), S(0)) → S(S(S(S(0)))) 0.00/0.72
save(S(x)) → dbl(0, save(x)) 0.00/0.72
save(0) → 0 0.00/0.72
dbl(0, y) → y

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

Converted CpxTRS to CDT
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0.00/0.72

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

dbl(S(0), S(0)) → S(S(S(S(0)))) 0.00/0.72
dbl(0, z0) → z0 0.00/0.72
save(S(z0)) → dbl(0, save(z0)) 0.00/0.72
save(0) → 0
Tuples:

SAVE(S(z0)) → c2(DBL(0, save(z0)), SAVE(z0))
S tuples:

SAVE(S(z0)) → c2(DBL(0, save(z0)), SAVE(z0))
K tuples:none
Defined Rule Symbols:

dbl, save

Defined Pair Symbols:

SAVE

Compound Symbols:

c2

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0.00/0.72

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

Removed 1 trailing tuple parts
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0.00/0.72

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

dbl(S(0), S(0)) → S(S(S(S(0)))) 0.00/0.72
dbl(0, z0) → z0 0.00/0.72
save(S(z0)) → dbl(0, save(z0)) 0.00/0.72
save(0) → 0
Tuples:

SAVE(S(z0)) → c2(SAVE(z0))
S tuples:

SAVE(S(z0)) → c2(SAVE(z0))
K tuples:none
Defined Rule Symbols:

dbl, save

Defined Pair Symbols:

SAVE

Compound Symbols:

c2

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0.00/0.72

(5) 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.

SAVE(S(z0)) → c2(SAVE(z0))
We considered the (Usable) Rules:none
And the Tuples:

SAVE(S(z0)) → c2(SAVE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.72

POL(S(x1)) = [1] + x1    0.00/0.72
POL(SAVE(x1)) = [3]x1    0.00/0.72
POL(c2(x1)) = x1   
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0.00/0.72

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

dbl(S(0), S(0)) → S(S(S(S(0)))) 0.00/0.72
dbl(0, z0) → z0 0.00/0.72
save(S(z0)) → dbl(0, save(z0)) 0.00/0.72
save(0) → 0
Tuples:

SAVE(S(z0)) → c2(SAVE(z0))
S tuples:none
K tuples:

SAVE(S(z0)) → c2(SAVE(z0))
Defined Rule Symbols:

dbl, save

Defined Pair Symbols:

SAVE

Compound Symbols:

c2

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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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0.00/0.73 EOF