The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

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

sqr(0) → 0
sqr(s(x)) → +(sqr(x), s(double(x)))
double(0) → 0
double(s(x)) → s(s(double(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
sqr(s(x)) → s(+(sqr(x), double(x)))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(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

(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))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))

We considered the (Usable) Rules:

sqr(0) → 0
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

And the Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+(x₁, x₂)) = [4] + [4]x₂   
POL(+'(x₁, x₂)) = [5]   
POL(0) = [5]   
POL(DOUBLE(x₁)) = [2]   
POL(SQR(x₁)) = [4]x₁   
POL(c1(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c2(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c4(x₁)) = x₁   
POL(c6(x₁)) = x₁   
POL(double(x₁)) = 0   
POL(s(x₁)) = [3] + x₁   
POL(sqr(x₁)) = [1] + [5]x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:

DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

K tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
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) 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
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

And the Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

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

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(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))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))

Defined Rule Symbols:

sqr, double, +

Defined Pair Symbols:

SQR, DOUBLE, +'

Compound Symbols:

c1, c2, c4, c6

(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
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

And the Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

sqr(0) → 0
sqr(s(z0)) → +(sqr(z0), s(double(z0)))
sqr(s(z0)) → s(+(sqr(z0), double(z0)))
double(0) → 0
double(s(z0)) → s(s(double(z0)))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

S tuples:none
K tuples:

SQR(s(z0)) → c1(+'(sqr(z0), s(double(z0))), SQR(z0), DOUBLE(z0))
SQR(s(z0)) → c2(+'(sqr(z0), double(z0)), SQR(z0), DOUBLE(z0))
DOUBLE(s(z0)) → c4(DOUBLE(z0))
+'(z0, s(z1)) → c6(+'(z0, z1))

Defined Rule Symbols:

sqr, double, +

Defined Pair Symbols:

SQR, DOUBLE, +'

Compound Symbols:

c1, c2, c4, c6

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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