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

(0) Obligation:

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

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

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))

S tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))

K tuples:none
Defined Rule Symbols:

double, +

Defined Pair Symbols:

DOUBLE, +'

Compound Symbols:

c1, c2, c4, c5

(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(z0) → c2(+'(z0, z0))

We considered the (Usable) Rules:none
And the Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))

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

POL(+'(x₁, x₂)) = 0   
POL(DOUBLE(x₁)) = [2]   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(s(x₁)) = 0   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))

S tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))

K tuples:

DOUBLE(z0) → c2(+'(z0, z0))

Defined Rule Symbols:

double, +

Defined Pair Symbols:

DOUBLE, +'

Compound Symbols:

c1, c2, c4, c5

(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.

DOUBLE(s(z0)) → c1(DOUBLE(z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))

We considered the (Usable) Rules:none
And the Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))

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

POL(+'(x₁, x₂)) = [2] + [2]x₁ + x₂   
POL(DOUBLE(x₁)) = [5] + [4]x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(s(x₁)) = [2] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
DOUBLE(z0) → c2(+'(z0, z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))

S tuples:none
K tuples:

DOUBLE(z0) → c2(+'(z0, z0))
DOUBLE(s(z0)) → c1(DOUBLE(z0))
+'(z0, s(z1)) → c4(+'(z0, z1))
+'(s(z0), z1) → c5(+'(z0, z1))

Defined Rule Symbols:

double, +

Defined Pair Symbols:

DOUBLE, +'

Compound Symbols:

c1, c2, c4, c5

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))