### (0) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(s(x), y) → +(x, s(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))
+(s(z0), z1) → +(z0, s(z1))
Tuples:

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

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

+

Defined Pair Symbols:

+'

Compound Symbols:

c, c1, c2

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

Removed 1 trailing nodes:

+'(0, z0) → c

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))
S tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))
K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c1, c2

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))
S tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

+'

Compound Symbols:

c1, c2

### (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), z1) → c2(+'(z0, s(z1)))
We considered the (Usable) Rules:none
And the Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+'(x1, x2)) = [4]x1
POL(c1(x1)) = x1
POL(c2(x1)) = x1
POL(s(x1)) = [4] + x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))
S tuples:none
K tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))
Defined Rule Symbols:none

Defined Pair Symbols:

+'

Compound Symbols:

c1, c2

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

The set S is empty