YES(O(1), O(n^1)) 0.00/0.69 YES(O(1), O(n^1)) 0.00/0.70 0.00/0.70 0.00/0.70 0.00/0.70 0.00/0.70 0.00/0.70 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.70 0.00/0.70 0.00/0.70
0.00/0.70 0.00/0.70 0.00/0.70
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0.00/0.70

(0) Obligation:

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

add0(S(x), x2) → +(S(0), add0(x2, x)) 0.00/0.70
add0(0, x2) → x2

The (relative) TRS S consists of the following rules:

+(x, S(0)) → S(x) 0.00/0.70
+(S(0), y) → S(y)

Rewrite Strategy: INNERMOST
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0.00/0.70

(1) CpxRelTrsToCDT (UPPER BOUND (ID) transformation)

Relative innermost TRS to CDT Problem.
0.00/0.70
0.00/0.70

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, S(0)) → S(z0) 0.00/0.70
+(S(0), z0) → S(z0) 0.00/0.70
add0(S(z0), z1) → +(S(0), add0(z1, z0)) 0.00/0.70
add0(0, z0) → z0
Tuples:

ADD0(S(z0), z1) → c2(+'(S(0), add0(z1, z0)), ADD0(z1, z0))
S tuples:

ADD0(S(z0), z1) → c2(+'(S(0), add0(z1, z0)), ADD0(z1, z0))
K tuples:none
Defined Rule Symbols:

add0, +

Defined Pair Symbols:

ADD0

Compound Symbols:

c2

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0.00/0.70

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

Removed 1 trailing tuple parts
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0.00/0.70

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, S(0)) → S(z0) 0.00/0.70
+(S(0), z0) → S(z0) 0.00/0.70
add0(S(z0), z1) → +(S(0), add0(z1, z0)) 0.00/0.70
add0(0, z0) → z0
Tuples:

ADD0(S(z0), z1) → c2(ADD0(z1, z0))
S tuples:

ADD0(S(z0), z1) → c2(ADD0(z1, z0))
K tuples:none
Defined Rule Symbols:

add0, +

Defined Pair Symbols:

ADD0

Compound Symbols:

c2

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0.00/0.70

(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.

ADD0(S(z0), z1) → c2(ADD0(z1, z0))
We considered the (Usable) Rules:none
And the Tuples:

ADD0(S(z0), z1) → c2(ADD0(z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.70

POL(ADD0(x1, x2)) = [5]x1 + [5]x2    0.00/0.70
POL(S(x1)) = [1] + x1    0.00/0.70
POL(c2(x1)) = x1   
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0.00/0.70

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(z0, S(0)) → S(z0) 0.00/0.70
+(S(0), z0) → S(z0) 0.00/0.70
add0(S(z0), z1) → +(S(0), add0(z1, z0)) 0.00/0.70
add0(0, z0) → z0
Tuples:

ADD0(S(z0), z1) → c2(ADD0(z1, z0))
S tuples:none
K tuples:

ADD0(S(z0), z1) → c2(ADD0(z1, z0))
Defined Rule Symbols:

add0, +

Defined Pair Symbols:

ADD0

Compound Symbols:

c2

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0.00/0.70

(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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0.00/0.72 EOF