### (0) Obligation:

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

g(S(x), y) → g(x, S(y))
f(y, S(x)) → f(S(y), x)
g(0, x2) → x2
f(x1, 0) → g(x1, 0)

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(S(z0), z1) → g(z0, S(z1))
g(0, z0) → z0
f(z0, S(z1)) → f(S(z0), z1)
f(z0, 0) → g(z0, 0)
Tuples:

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

G(S(z0), z1) → c(G(z0, S(z1)))
G(0, z0) → c1
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
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 1 trailing nodes:

G(0, z0) → c1

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(S(z0), z1) → g(z0, S(z1))
g(0, z0) → z0
f(z0, S(z1)) → f(S(z0), z1)
f(z0, 0) → g(z0, 0)
Tuples:

G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
S tuples:

G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
K tuples:none
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G, F

Compound Symbols:

c, c2, c3

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

g(S(z0), z1) → g(z0, S(z1))
g(0, z0) → z0
f(z0, S(z1)) → f(S(z0), z1)
f(z0, 0) → g(z0, 0)

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
S tuples:

G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G, F

Compound Symbols:

c, 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.

G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
We considered the (Usable) Rules:none
And the Tuples:

G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [4]
POL(F(x1, x2)) = [4] + [2]x1 + [3]x2
POL(G(x1, x2)) = [5] + [2]x1
POL(S(x1)) = [4] + x1
POL(c(x1)) = x1
POL(c2(x1)) = x1
POL(c3(x1)) = x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
S tuples:none
K tuples:

G(S(z0), z1) → c(G(z0, S(z1)))
F(z0, S(z1)) → c2(F(S(z0), z1))
F(z0, 0) → c3(G(z0, 0))
Defined Rule Symbols:none

Defined Pair Symbols:

G, F

Compound Symbols:

c, c2, c3

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

The set S is empty