### (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