### (0) Obligation:

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

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

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
-(0, z0) → 0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)
Tuples:

+'(0, z0) → c
+'(s(z0), z1) → c1(+'(z0, z1))
-'(0, z0) → c2
-'(z0, 0) → c3
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:

+'(0, z0) → c
+'(s(z0), z1) → c1(+'(z0, z1))
-'(0, z0) → c2
-'(z0, 0) → c3
-'(s(z0), s(z1)) → c4(-'(z0, z1))
K tuples:none
Defined Rule Symbols:

+, -

Defined Pair Symbols:

+', -'

Compound Symbols:

c, c1, c2, c3, c4

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

Removed 3 trailing nodes:

+'(0, z0) → c
-'(0, z0) → c2
-'(z0, 0) → c3

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

+(0, z0) → z0
+(s(z0), z1) → s(+(z0, z1))
-(0, z0) → 0
-(z0, 0) → z0
-(s(z0), s(z1)) → -(z0, z1)
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
K tuples:none
Defined Rule Symbols:

+, -

Defined Pair Symbols:

+', -'

Compound Symbols:

c1, c4

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

+', -'

Compound Symbols:

c1, c4

### (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.

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+'(x1, x2)) = [2]x1
POL(-'(x1, x2)) = [2]x2
POL(c1(x1)) = x1
POL(c4(x1)) = x1
POL(s(x1)) = [3] + x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
S tuples:none
K tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
-'(s(z0), s(z1)) → c4(-'(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

+', -'

Compound Symbols:

c1, c4

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

The set S is empty