YES(O(1), O(n^1)) 0.00/0.87 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 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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↳4 CdtProblem
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↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
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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 SIsEmptyProof (BOTH BOUNDS(ID, ID))
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↳10 BOUNDS(O(1), O(1))
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0.00/0.89 0.00/0.89
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0.00/0.89

(0) Obligation:

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

double(0) → 0 0.00/0.89
double(s(x)) → s(s(double(x))) 0.00/0.89
half(0) → 0 0.00/0.89
half(s(0)) → 0 0.00/0.89
half(s(s(x))) → s(half(x)) 0.00/0.89
-(x, 0) → x 0.00/0.89
-(s(x), s(y)) → -(x, y) 0.00/0.89
if(0, y, z) → y 0.00/0.89
if(s(x), y, z) → z 0.00/0.89
half(double(x)) → x

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:

double(0) → 0 0.00/0.89
double(s(z0)) → s(s(double(z0))) 0.00/0.89
half(0) → 0 0.00/0.89
half(s(0)) → 0 0.00/0.89
half(s(s(z0))) → s(half(z0)) 0.00/0.89
half(double(z0)) → z0 0.00/0.89
-(z0, 0) → z0 0.00/0.89
-(s(z0), s(z1)) → -(z0, z1) 0.00/0.89
if(0, z0, z1) → z0 0.00/0.89
if(s(z0), z1, z2) → z2
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.89
HALF(s(s(z0))) → c4(HALF(z0)) 0.00/0.89
-'(s(z0), s(z1)) → c7(-'(z0, z1))
S tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.89
HALF(s(s(z0))) → c4(HALF(z0)) 0.00/0.89
-'(s(z0), s(z1)) → c7(-'(z0, z1))
K tuples:none
Defined Rule Symbols:

double, half, -, if

Defined Pair Symbols:

DOUBLE, HALF, -'

Compound Symbols:

c1, c4, c7

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(3) 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.

DOUBLE(s(z0)) → c1(DOUBLE(z0))
We considered the (Usable) Rules:none
And the Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.89
HALF(s(s(z0))) → c4(HALF(z0)) 0.00/0.89
-'(s(z0), s(z1)) → c7(-'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.89

POL(-'(x1, x2)) = 0    0.00/0.89
POL(DOUBLE(x1)) = [2]x1    0.00/0.89
POL(HALF(x1)) = 0    0.00/0.89
POL(c1(x1)) = x1    0.00/0.89
POL(c4(x1)) = x1    0.00/0.89
POL(c7(x1)) = x1    0.00/0.89
POL(s(x1)) = [1] + x1   
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(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

double(0) → 0 0.00/0.89
double(s(z0)) → s(s(double(z0))) 0.00/0.89
half(0) → 0 0.00/0.89
half(s(0)) → 0 0.00/0.89
half(s(s(z0))) → s(half(z0)) 0.00/0.89
half(double(z0)) → z0 0.00/0.89
-(z0, 0) → z0 0.00/0.89
-(s(z0), s(z1)) → -(z0, z1) 0.00/0.89
if(0, z0, z1) → z0 0.00/0.89
if(s(z0), z1, z2) → z2
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.89
HALF(s(s(z0))) → c4(HALF(z0)) 0.00/0.89
-'(s(z0), s(z1)) → c7(-'(z0, z1))
S tuples:

HALF(s(s(z0))) → c4(HALF(z0)) 0.00/0.89
-'(s(z0), s(z1)) → c7(-'(z0, z1))
K tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
Defined Rule Symbols:

double, half, -, if

Defined Pair Symbols:

DOUBLE, HALF, -'

Compound Symbols:

c1, c4, c7

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(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.

HALF(s(s(z0))) → c4(HALF(z0))
We considered the (Usable) Rules:none
And the Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.89
HALF(s(s(z0))) → c4(HALF(z0)) 0.00/0.89
-'(s(z0), s(z1)) → c7(-'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.89

POL(-'(x1, x2)) = 0    0.00/0.89
POL(DOUBLE(x1)) = [5]x1    0.00/0.89
POL(HALF(x1)) = [4]x1    0.00/0.89
POL(c1(x1)) = x1    0.00/0.89
POL(c4(x1)) = x1    0.00/0.89
POL(c7(x1)) = x1    0.00/0.89
POL(s(x1)) = [4] + x1   
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(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

double(0) → 0 0.00/0.89
double(s(z0)) → s(s(double(z0))) 0.00/0.89
half(0) → 0 0.00/0.89
half(s(0)) → 0 0.00/0.89
half(s(s(z0))) → s(half(z0)) 0.00/0.89
half(double(z0)) → z0 0.00/0.89
-(z0, 0) → z0 0.00/0.89
-(s(z0), s(z1)) → -(z0, z1) 0.00/0.89
if(0, z0, z1) → z0 0.00/0.89
if(s(z0), z1, z2) → z2
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.89
HALF(s(s(z0))) → c4(HALF(z0)) 0.00/0.89
-'(s(z0), s(z1)) → c7(-'(z0, z1))
S tuples:

-'(s(z0), s(z1)) → c7(-'(z0, z1))
K tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.89
HALF(s(s(z0))) → c4(HALF(z0))
Defined Rule Symbols:

double, half, -, if

Defined Pair Symbols:

DOUBLE, HALF, -'

Compound Symbols:

c1, c4, c7

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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.

-'(s(z0), s(z1)) → c7(-'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.89
HALF(s(s(z0))) → c4(HALF(z0)) 0.00/0.89
-'(s(z0), s(z1)) → c7(-'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.89

POL(-'(x1, x2)) = x1    0.00/0.89
POL(DOUBLE(x1)) = [5]x1    0.00/0.89
POL(HALF(x1)) = [5]x1    0.00/0.89
POL(c1(x1)) = x1    0.00/0.89
POL(c4(x1)) = x1    0.00/0.89
POL(c7(x1)) = x1    0.00/0.89
POL(s(x1)) = [2] + x1   
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(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

double(0) → 0 0.00/0.89
double(s(z0)) → s(s(double(z0))) 0.00/0.89
half(0) → 0 0.00/0.89
half(s(0)) → 0 0.00/0.89
half(s(s(z0))) → s(half(z0)) 0.00/0.89
half(double(z0)) → z0 0.00/0.89
-(z0, 0) → z0 0.00/0.89
-(s(z0), s(z1)) → -(z0, z1) 0.00/0.89
if(0, z0, z1) → z0 0.00/0.89
if(s(z0), z1, z2) → z2
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.89
HALF(s(s(z0))) → c4(HALF(z0)) 0.00/0.89
-'(s(z0), s(z1)) → c7(-'(z0, z1))
S tuples:none
K tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.89
HALF(s(s(z0))) → c4(HALF(z0)) 0.00/0.89
-'(s(z0), s(z1)) → c7(-'(z0, z1))
Defined Rule Symbols:

double, half, -, if

Defined Pair Symbols:

DOUBLE, HALF, -'

Compound Symbols:

c1, c4, c7

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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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0.00/0.96 EOF