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

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

double(0) → 0
double(s(z0)) → s(s(double(z0)))
half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))
half(double(z0)) → z0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)
if(0, z0, z1) → z0
if(s(z0), z1, z2) → z2

Tuples:

DOUBLE(0) → c
DOUBLE(s(z0)) → c1(DOUBLE(z0))
HALF(0) → c2
HALF(s(0)) → c3
HALF(s(s(z0))) → c4(HALF(z0))
HALF(double(z0)) → c5
-'(z0, 0) → c6
-'(s(z0), s(z1)) → c7(-'(z0, z1))
IF(0, z0, z1) → c8
IF(s(z0), z1, z2) → c9

S tuples:

DOUBLE(0) → c
DOUBLE(s(z0)) → c1(DOUBLE(z0))
HALF(0) → c2
HALF(s(0)) → c3
HALF(s(s(z0))) → c4(HALF(z0))
HALF(double(z0)) → c5
-'(z0, 0) → c6
-'(s(z0), s(z1)) → c7(-'(z0, z1))
IF(0, z0, z1) → c8
IF(s(z0), z1, z2) → c9

K tuples:none
Defined Rule Symbols:

double, half, -, if

Defined Pair Symbols:

DOUBLE, HALF, -', IF

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 7 trailing nodes:

IF(s(z0), z1, z2) → c9
DOUBLE(0) → c
HALF(0) → c2
HALF(s(0)) → c3
IF(0, z0, z1) → c8
HALF(double(z0)) → c5
-'(z0, 0) → c6

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

double(0) → 0
double(s(z0)) → s(s(double(z0)))
half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))
half(double(z0)) → z0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)
if(0, z0, z1) → z0
if(s(z0), z1, z2) → z2

Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
HALF(s(s(z0))) → c4(HALF(z0))
-'(s(z0), s(z1)) → c7(-'(z0, z1))

S tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
HALF(s(s(z0))) → c4(HALF(z0))
-'(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

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

double(0) → 0
double(s(z0)) → s(s(double(z0)))
half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))
half(double(z0)) → z0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)
if(0, z0, z1) → z0
if(s(z0), z1, z2) → z2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
HALF(s(s(z0))) → c4(HALF(z0))
-'(s(z0), s(z1)) → c7(-'(z0, z1))

S tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
HALF(s(s(z0))) → c4(HALF(z0))
-'(s(z0), s(z1)) → c7(-'(z0, z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

DOUBLE, HALF, -'

Compound Symbols:

c1, c4, c7

(7) CdtRuleRemovalProof (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))
-'(s(z0), s(z1)) → c7(-'(z0, z1))

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

DOUBLE(s(z0)) → c1(DOUBLE(z0))
HALF(s(s(z0))) → c4(HALF(z0))
-'(s(z0), s(z1)) → c7(-'(z0, z1))

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

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
HALF(s(s(z0))) → c4(HALF(z0))
-'(s(z0), s(z1)) → c7(-'(z0, z1))

S tuples:

HALF(s(s(z0))) → c4(HALF(z0))

K tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
-'(s(z0), s(z1)) → c7(-'(z0, z1))

Defined Rule Symbols:none

Defined Pair Symbols:

DOUBLE, HALF, -'

Compound Symbols:

c1, c4, c7

(9) CdtRuleRemovalProof (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))
HALF(s(s(z0))) → c4(HALF(z0))
-'(s(z0), s(z1)) → c7(-'(z0, z1))

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

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

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
HALF(s(s(z0))) → c4(HALF(z0))
-'(s(z0), s(z1)) → c7(-'(z0, z1))

S tuples:none
K tuples:

DOUBLE(s(z0)) → c1(DOUBLE(z0))
-'(s(z0), s(z1)) → c7(-'(z0, z1))
HALF(s(s(z0))) → c4(HALF(z0))

Defined Rule Symbols:none

Defined Pair Symbols:

DOUBLE, HALF, -'

Compound Symbols:

c1, c4, c7

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

The set S is empty

(12) BOUNDS(O(1), O(1))