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

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(z0, s(z1)) → c1(G(f(z0, z1), 0))
G(s(z0), z1) → c2(G(f(z0, z1), 0))
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))

S tuples:

G(z0, s(z1)) → c1(G(f(z0, z1), 0))
G(s(z0), z1) → c2(G(f(z0, z1), 0))
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))

K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c1, c2, c3

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

G(z0, s(z1)) → c1(G(f(z0, z1), 0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(s(z0), z1) → c2(G(f(z0, z1), 0))
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))

S tuples:

G(s(z0), z1) → c2(G(f(z0, z1), 0))
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))

K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c2, c3

(5) CdtPolyRedPairProof (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) → c2(G(f(z0, z1), 0))
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))

We considered the (Usable) Rules:none
And the Tuples:

G(s(z0), z1) → c2(G(f(z0, z1), 0))
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(G(x₁, x₂)) = [2] + [4]x₁ + [4]x₂   
POL(c2(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(f(x₁, x₂)) = [2] + x₁ + x₂   
POL(s(x₁)) = [3] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(s(z0), z1) → c2(G(f(z0, z1), 0))
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))

S tuples:none
K tuples:

G(s(z0), z1) → c2(G(f(z0, z1), 0))
G(f(z0, z1), 0) → c3(G(z0, 0), G(z1, 0))

Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c2, c3

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))