The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

(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

(10) BOUNDS(O(1), O(1))