YES(O(1), O(n^1)) 0.00/0.75 YES(O(1), O(n^1)) 0.00/0.76 0.00/0.76 0.00/0.76 0.00/0.76 0.00/0.76 0.00/0.76 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 0.00/0.76 0.00/0.76 0.00/0.76
0.00/0.76 0.00/0.76 0.00/0.76
0.00/0.76
0.00/0.76

(0) Obligation:

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

double(0) → 0 0.00/0.76
double(s(x)) → s(s(double(x))) 0.00/0.76
+(x, 0) → x 0.00/0.76
+(x, s(y)) → s(+(x, y)) 0.00/0.76
+(s(x), y) → s(+(x, y)) 0.00/0.76
double(x) → +(x, x)

Rewrite Strategy: INNERMOST
0.00/0.76
0.00/0.76

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT
0.00/0.76
0.00/0.76

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

double(0) → 0 0.00/0.76
double(s(z0)) → s(s(double(z0))) 0.00/0.76
double(z0) → +(z0, z0) 0.00/0.76
+(z0, 0) → z0 0.00/0.76
+(z0, s(z1)) → s(+(z0, z1)) 0.00/0.76
+(s(z0), z1) → s(+(z0, z1))
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.76
DOUBLE(z0) → c2(+'(z0, z0)) 0.00/0.76
+'(z0, s(z1)) → c4(+'(z0, z1)) 0.00/0.76
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.76
DOUBLE(z0) → c2(+'(z0, z0)) 0.00/0.76
+'(z0, s(z1)) → c4(+'(z0, z1)) 0.00/0.76
+'(s(z0), z1) → c5(+'(z0, z1))
K tuples:none
Defined Rule Symbols:

double, +

Defined Pair Symbols:

DOUBLE, +'

Compound Symbols:

c1, c2, c4, c5

0.00/0.76
0.00/0.76

(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)) 0.00/0.76
DOUBLE(z0) → c2(+'(z0, z0)) 0.00/0.76
+'(z0, s(z1)) → c4(+'(z0, z1)) 0.00/0.76
+'(s(z0), z1) → c5(+'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.76
DOUBLE(z0) → c2(+'(z0, z0)) 0.00/0.76
+'(z0, s(z1)) → c4(+'(z0, z1)) 0.00/0.76
+'(s(z0), z1) → c5(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 0.00/0.76

POL(+'(x1, x2)) = [1] + x1 + x2    0.00/0.76
POL(DOUBLE(x1)) = [2] + [5]x1    0.00/0.76
POL(c1(x1)) = x1    0.00/0.76
POL(c2(x1)) = x1    0.00/0.76
POL(c4(x1)) = x1    0.00/0.76
POL(c5(x1)) = x1    0.00/0.76
POL(s(x1)) = [3] + x1   
0.00/0.76
0.00/0.76

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

double(0) → 0 0.00/0.76
double(s(z0)) → s(s(double(z0))) 0.00/0.76
double(z0) → +(z0, z0) 0.00/0.76
+(z0, 0) → z0 0.00/0.76
+(z0, s(z1)) → s(+(z0, z1)) 0.00/0.76
+(s(z0), z1) → s(+(z0, z1))
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.76
DOUBLE(z0) → c2(+'(z0, z0)) 0.00/0.76
+'(z0, s(z1)) → c4(+'(z0, z1)) 0.00/0.76
+'(s(z0), z1) → c5(+'(z0, z1))
S tuples:none
K tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0)) 0.00/0.76
DOUBLE(z0) → c2(+'(z0, z0)) 0.00/0.76
+'(z0, s(z1)) → c4(+'(z0, z1)) 0.00/0.76
+'(s(z0), z1) → c5(+'(z0, z1))
Defined Rule Symbols:

double, +

Defined Pair Symbols:

DOUBLE, +'

Compound Symbols:

c1, c2, c4, c5

0.00/0.76
0.00/0.76

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

The set S is empty
0.00/0.76
0.00/0.76

(6) BOUNDS(O(1), O(1))

0.00/0.76
0.00/0.76
0.00/0.78 EOF