### (0) Obligation:

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

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

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(g(z0), s(0)) → c(F(g(z0), g(z0)), G(z0), G(z0))
G(s(z0)) → c1(G(z0))
G(0) → c2
S tuples:

F(g(z0), s(0)) → c(F(g(z0), g(z0)), G(z0), G(z0))
G(s(z0)) → c1(G(z0))
G(0) → c2
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1, c2

### (3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

F(g(z0), s(0)) → c(F(g(z0), g(z0)), G(z0), G(z0))
Removed 1 trailing nodes:

G(0) → c2

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

G(s(z0)) → c1(G(z0))
S tuples:

G(s(z0)) → c1(G(z0))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

G

Compound Symbols:

c1

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(g(z0), s(0)) → f(g(z0), g(z0))
g(s(z0)) → s(g(z0))
g(0) → 0

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(s(z0)) → c1(G(z0))
S tuples:

G(s(z0)) → c1(G(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c1

### (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.

G(s(z0)) → c1(G(z0))
We considered the (Usable) Rules:none
And the Tuples:

G(s(z0)) → c1(G(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(G(x1)) = [5]x1
POL(c1(x1)) = x1
POL(s(x1)) = [1] + x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(s(z0)) → c1(G(z0))
S tuples:none
K tuples:

G(s(z0)) → c1(G(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c1

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

The set S is empty