### (0) Obligation:

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

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

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

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

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1, c2, c3, c4

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

Removed 3 trailing nodes:

F(0) → c
F(s(0)) → c1
F(s(0)) → c2(F(0))

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))
S tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))
S tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c3, 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.

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))
We considered the (Usable) Rules:none
And the Tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(+(x1, x2)) = [3] + x1 + x2
POL(0) = [1]
POL(F(x1)) = [1] + [3]x1
POL(c3(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(s(x1)) = x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))
S tuples:none
K tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

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

The set S is empty