### (0) Obligation:

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(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:

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

H(z0, z1) → c(F(z0, s(z0), z1))
F(z0, z1, g(z0, z1)) → c1(H(0, g(z0, z1)), G(z0, z1))
G(0, z0) → c2
G(z0, s(z1)) → c3(G(z0, z1))
S tuples:

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

h, f, g

Defined Pair Symbols:

H, F, G

Compound Symbols:

c, c1, c2, c3

### (3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

F(z0, z1, g(z0, z1)) → c1(H(0, g(z0, z1)), G(z0, z1))
Removed 2 trailing nodes:

H(z0, z1) → c(F(z0, s(z0), z1))
G(0, z0) → c2

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

G(z0, s(z1)) → c3(G(z0, z1))
S tuples:

G(z0, s(z1)) → c3(G(z0, z1))
K tuples:none
Defined Rule Symbols:

h, f, g

Defined Pair Symbols:

G

Compound Symbols:

c3

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

h(z0, z1) → f(z0, s(z0), z1)
f(z0, z1, g(z0, z1)) → h(0, g(z0, z1))
g(0, z0) → 0
g(z0, s(z1)) → g(z0, z1)

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(z0, s(z1)) → c3(G(z0, z1))
S tuples:

G(z0, s(z1)) → c3(G(z0, z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c3

### (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(z0, s(z1)) → c3(G(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

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

POL(G(x1, x2)) = [5]x2
POL(c3(x1)) = x1
POL(s(x1)) = [1] + x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(z0, s(z1)) → c3(G(z0, z1))
S tuples:none
K tuples:

G(z0, s(z1)) → c3(G(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c3

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

The set S is empty