### (0) Obligation:

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

f(0, y) → 0
f(s(x), y) → f(f(x, y), y)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(0, z0) → c
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:

F(0, z0) → c
F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

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

Removed 1 trailing nodes:

F(0, z0) → c

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

### (5) 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.

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
We considered the (Usable) Rules:

f(0, z0) → 0
f(s(z0), z1) → f(f(z0, z1), z1)
And the Tuples:

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

POL(0) = 0
POL(F(x1, x2)) = [3] + [4]x1
POL(c1(x1, x2)) = x1 + x2
POL(f(x1, x2)) = 0
POL(s(x1)) = [4] + x1

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
S tuples:none
K tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

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

The set S is empty