### (0) Obligation:

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

g(x, y) → x
g(x, y) → y
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(z0, z1) → z0
g(z0, z1) → z1
f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))
Tuples:

G(z0, z1) → c
G(z0, z1) → c1
F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
S tuples:

G(z0, z1) → c
G(z0, z1) → c1
F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c, c1, c2, c3

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

Removed 2 trailing nodes:

G(z0, z1) → c
G(z0, z1) → c1

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(z0, z1) → z0
g(z0, z1) → z1
f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))
Tuples:

F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
S tuples:

F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:

g, f

Defined Pair Symbols:

F

Compound Symbols:

c2, c3

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

g(z0, z1) → z0
g(z0, z1) → z1
f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
S tuples:

F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c2, 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.

F(z0, z1, s(z2)) → c3(F(0, 1, z2))
We considered the (Usable) Rules:none
And the Tuples:

F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [4]
POL(1) = 0
POL(F(x1, x2, x3)) = [4]x3
POL(c2(x1)) = x1
POL(c3(x1)) = x1
POL(s(x1)) = [4] + x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
S tuples:

F(0, 1, z0) → c2(F(s(z0), z0, z0))
K tuples:

F(z0, z1, s(z2)) → c3(F(0, 1, z2))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c2, c3

### (9) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

F(0, 1, z0) → c2(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c3(F(0, 1, z2))
Now S is empty