YES(O(1), O(n^2)) 5.17/1.79 YES(O(1), O(n^2)) 5.55/1.82 5.55/1.82 5.55/1.82 5.55/1.82 5.55/1.82 5.55/1.82 Runtime Complexity (innermost) proof of /export/starexec/sandbox/benchmark/theBenchmark.xml.xml 5.55/1.82 5.55/1.82 5.55/1.82
5.55/1.82 5.55/1.82 5.55/1.82
5.55/1.82
5.55/1.82

(0) Obligation:

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

sqr(0) → 0 5.55/1.82
sqr(s(x)) → +(sqr(x), s(double(x))) 5.55/1.82
double(0) → 0 5.55/1.82
double(s(x)) → s(s(double(x))) 5.55/1.82
+(x, 0) → x 5.55/1.82
+(x, s(y)) → s(+(x, y)) 5.55/1.82
sqr(s(x)) → s(+(sqr(x), double(x)))

Rewrite Strategy: INNERMOST
5.55/1.82
5.55/1.82

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

Converted CpxTRS to CDT
5.55/1.82
5.55/1.82

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0 5.55/1.82
sqr(s(z0)) → +(sqr(z0), s(double(z0))) 5.55/1.82
sqr(s(z0)) → s(+(sqr(z0), double(z0))) 5.55/1.82
double(0) → 0 5.55/1.82
double(s(z0)) → s(s(double(z0))) 5.55/1.82
+(z0, 0) → z0 5.55/1.82
+(z0, s(z1)) → s(+(z0, z1))
Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.82
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0)) 5.55/1.82
DOUBLE(s(z0)) → c4(DOUBLE(z0)) 5.55/1.82
+'(z0, s(z1)) → c6(+'(z0, z1))
S tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.82
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0)) 5.55/1.82
DOUBLE(s(z0)) → c4(DOUBLE(z0)) 5.55/1.82
+'(z0, s(z1)) → c6(+'(z0, z1))
K tuples:none
Defined Rule Symbols:

sqr, double, +

Defined Pair Symbols:

SQR, DOUBLE, +'

Compound Symbols:

c1, c2, c4, c6

5.55/1.82
5.55/1.82

(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.

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.82
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
We considered the (Usable) Rules:

sqr(0) → 0 5.55/1.82
sqr(s(z0)) → +(sqr(z0), s(double(z0))) 5.55/1.82
sqr(s(z0)) → s(+(sqr(z0), double(z0))) 5.55/1.82
double(0) → 0 5.55/1.82
double(s(z0)) → s(s(double(z0))) 5.55/1.82
+(z0, 0) → z0 5.55/1.82
+(z0, s(z1)) → s(+(z0, z1))
And the Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.82
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0)) 5.55/1.82
DOUBLE(s(z0)) → c4(DOUBLE(z0)) 5.55/1.82
+'(z0, s(z1)) → c6(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.55/1.82

POL(+(x1, x2)) = [5] + [3]x2    5.55/1.82
POL(+'(x1, x2)) = 0    5.55/1.82
POL(0) = [1]    5.55/1.82
POL(DOUBLE(x1)) = 0    5.55/1.82
POL(SQR(x1)) = [4]x1    5.55/1.82
POL(c1(x1, x2, x3)) = x1 + x2 + x3    5.55/1.82
POL(c2(x1, x2, x3)) = x1 + x2 + x3    5.55/1.82
POL(c4(x1)) = x1    5.55/1.82
POL(c6(x1)) = x1    5.55/1.82
POL(double(x1)) = 0    5.55/1.82
POL(s(x1)) = [1] + x1    5.55/1.82
POL(sqr(x1)) = [4] + [4]x1   
5.55/1.82
5.55/1.82

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0 5.55/1.82
sqr(s(z0)) → +(sqr(z0), s(double(z0))) 5.55/1.82
sqr(s(z0)) → s(+(sqr(z0), double(z0))) 5.55/1.82
double(0) → 0 5.55/1.82
double(s(z0)) → s(s(double(z0))) 5.55/1.83
+(z0, 0) → z0 5.55/1.83
+(z0, s(z1)) → s(+(z0, z1))
Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.83
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0)) 5.55/1.83
DOUBLE(s(z0)) → c4(DOUBLE(z0)) 5.55/1.83
+'(z0, s(z1)) → c6(+'(z0, z1))
S tuples:

DOUBLE(s(z0)) → c4(DOUBLE(z0)) 5.55/1.83
+'(z0, s(z1)) → c6(+'(z0, z1))
K tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.83
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
Defined Rule Symbols:

sqr, double, +

Defined Pair Symbols:

SQR, DOUBLE, +'

Compound Symbols:

c1, c2, c4, c6

5.55/1.83
5.55/1.83

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

DOUBLE(s(z0)) → c4(DOUBLE(z0))
We considered the (Usable) Rules:

sqr(0) → 0 5.55/1.83
sqr(s(z0)) → +(sqr(z0), s(double(z0))) 5.55/1.83
sqr(s(z0)) → s(+(sqr(z0), double(z0))) 5.55/1.83
double(0) → 0 5.55/1.83
double(s(z0)) → s(s(double(z0))) 5.55/1.83
+(z0, 0) → z0 5.55/1.83
+(z0, s(z1)) → s(+(z0, z1))
And the Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.83
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0)) 5.55/1.83
DOUBLE(s(z0)) → c4(DOUBLE(z0)) 5.55/1.83
+'(z0, s(z1)) → c6(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.55/1.83

POL(+(x1, x2)) = [3]    5.55/1.83
POL(+'(x1, x2)) = 0    5.55/1.83
POL(0) = 0    5.55/1.83
POL(DOUBLE(x1)) = x1    5.55/1.83
POL(SQR(x1)) = [2]x12    5.55/1.83
POL(c1(x1, x2, x3)) = x1 + x2 + x3    5.55/1.83
POL(c2(x1, x2, x3)) = x1 + x2 + x3    5.55/1.83
POL(c4(x1)) = x1    5.55/1.83
POL(c6(x1)) = x1    5.55/1.83
POL(double(x1)) = [2] + x1 + x12    5.55/1.83
POL(s(x1)) = [3] + x1    5.55/1.83
POL(sqr(x1)) = 0   
5.55/1.83
5.55/1.83

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0 5.55/1.83
sqr(s(z0)) → +(sqr(z0), s(double(z0))) 5.55/1.83
sqr(s(z0)) → s(+(sqr(z0), double(z0))) 5.55/1.83
double(0) → 0 5.55/1.83
double(s(z0)) → s(s(double(z0))) 5.55/1.83
+(z0, 0) → z0 5.55/1.83
+(z0, s(z1)) → s(+(z0, z1))
Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.83
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0)) 5.55/1.83
DOUBLE(s(z0)) → c4(DOUBLE(z0)) 5.55/1.83
+'(z0, s(z1)) → c6(+'(z0, z1))
S tuples:

+'(z0, s(z1)) → c6(+'(z0, z1))
K tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.83
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0)) 5.55/1.83
DOUBLE(s(z0)) → c4(DOUBLE(z0))
Defined Rule Symbols:

sqr, double, +

Defined Pair Symbols:

SQR, DOUBLE, +'

Compound Symbols:

c1, c2, c4, c6

5.55/1.83
5.55/1.83

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

+'(z0, s(z1)) → c6(+'(z0, z1))
We considered the (Usable) Rules:

sqr(0) → 0 5.55/1.83
sqr(s(z0)) → +(sqr(z0), s(double(z0))) 5.55/1.83
sqr(s(z0)) → s(+(sqr(z0), double(z0))) 5.55/1.83
double(0) → 0 5.55/1.83
double(s(z0)) → s(s(double(z0))) 5.55/1.83
+(z0, 0) → z0 5.55/1.83
+(z0, s(z1)) → s(+(z0, z1))
And the Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.83
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0)) 5.55/1.83
DOUBLE(s(z0)) → c4(DOUBLE(z0)) 5.55/1.83
+'(z0, s(z1)) → c6(+'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation : 5.55/1.83

POL(+(x1, x2)) = [3]    5.55/1.83
POL(+'(x1, x2)) = x2    5.55/1.83
POL(0) = 0    5.55/1.83
POL(DOUBLE(x1)) = 0    5.55/1.83
POL(SQR(x1)) = x12    5.55/1.83
POL(c1(x1, x2, x3)) = x1 + x2 + x3    5.55/1.83
POL(c2(x1, x2, x3)) = x1 + x2 + x3    5.55/1.83
POL(c4(x1)) = x1    5.55/1.83
POL(c6(x1)) = x1    5.55/1.83
POL(double(x1)) = [2]x1    5.55/1.83
POL(s(x1)) = [1] + x1    5.55/1.83
POL(sqr(x1)) = 0   
5.55/1.83
5.55/1.83

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0 5.55/1.83
sqr(s(z0)) → +(sqr(z0), s(double(z0))) 5.55/1.83
sqr(s(z0)) → s(+(sqr(z0), double(z0))) 5.55/1.83
double(0) → 0 5.55/1.83
double(s(z0)) → s(s(double(z0))) 5.55/1.83
+(z0, 0) → z0 5.55/1.83
+(z0, s(z1)) → s(+(z0, z1))
Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.83
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0)) 5.55/1.83
DOUBLE(s(z0)) → c4(DOUBLE(z0)) 5.55/1.83
+'(z0, s(z1)) → c6(+'(z0, z1))
S tuples:none
K tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0)) 5.55/1.83
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0)) 5.55/1.83
DOUBLE(s(z0)) → c4(DOUBLE(z0)) 5.55/1.83
+'(z0, s(z1)) → c6(+'(z0, z1))
Defined Rule Symbols:

sqr, double, +

Defined Pair Symbols:

SQR, DOUBLE, +'

Compound Symbols:

c1, c2, c4, c6

5.55/1.83
5.55/1.83

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

The set S is empty
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5.55/1.83

(10) BOUNDS(O(1), O(1))

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5.55/1.83
5.55/1.86 EOF