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:

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

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(g(z0), s(0), z1) → c(F(g(s(0)), z1, g(z0)), G(s(0)), G(z0))
G(s(z0)) → c1(G(z0))

S tuples:

F(g(z0), s(0), z1) → c(F(g(s(0)), z1, g(z0)), G(s(0)), G(z0))
G(s(z0)) → c1(G(z0))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1

(3) 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.

F(g(z0), s(0), z1) → c(F(g(s(0)), z1, g(z0)), G(s(0)), G(z0))

We considered the (Usable) Rules:

g(s(z0)) → s(g(z0))
g(0) → 0

And the Tuples:

F(g(z0), s(0), z1) → c(F(g(s(0)), z1, g(z0)), G(s(0)), G(z0))
G(s(z0)) → c1(G(z0))

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

POL(0) = [3]   
POL(F(x₁, x₂, x₃)) = [3]x₂ + [3]x₃   
POL(G(x₁)) = [2]   
POL(c(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(c1(x₁)) = x₁   
POL(g(x₁)) = [1]   
POL(s(x₁)) = x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(g(z0), s(0), z1) → c(F(g(s(0)), z1, g(z0)), G(s(0)), G(z0))
G(s(z0)) → c1(G(z0))

S tuples:

G(s(z0)) → c1(G(z0))

K tuples:

F(g(z0), s(0), z1) → c(F(g(s(0)), z1, g(z0)), G(s(0)), G(z0))

Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1

(5) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace F(g(z0), s(0), z1) → c(F(g(s(0)), z1, g(z0)), G(s(0)), G(z0)) by

F(g(x0), s(0), x1) → c(F(s(g(0)), x1, g(x0)), G(s(0)), G(x0))
F(g(x0), s(0), x1) → c(G(x0))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(s(z0)) → c1(G(z0))
F(g(x0), s(0), x1) → c(F(s(g(0)), x1, g(x0)), G(s(0)), G(x0))
F(g(x0), s(0), x1) → c(G(x0))

S tuples:

G(s(z0)) → c1(G(z0))

K tuples:

F(g(z0), s(0), z1) → c(F(g(s(0)), z1, g(z0)), G(s(0)), G(z0))

Defined Rule Symbols:

f, g

Defined Pair Symbols:

G, F

Compound Symbols:

c1, c, c

(7) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 2 leading nodes:

F(g(x0), s(0), x1) → c(F(s(g(0)), x1, g(x0)), G(s(0)), G(x0))
F(g(x0), s(0), x1) → c(G(x0))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(s(z0)) → c1(G(z0))

S tuples:

G(s(z0)) → c1(G(z0))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

G

Compound Symbols:

c1

(9) 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)) → c1(G(z0))

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

G(s(z0)) → c1(G(z0))

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

POL(G(x₁)) = [5]x₁   
POL(c1(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(s(z0)) → c1(G(z0))

S tuples:none
K tuples:

G(s(z0)) → c1(G(z0))

Defined Rule Symbols:

f, g

Defined Pair Symbols:

G

Compound Symbols:

c1

(11) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(12) BOUNDS(O(1), O(1))