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(f(x)) → f(x)
g(0) → g(f(0))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(z0)
g(0) → g(f(0))

Tuples:

F(f(z0)) → c(F(z0))
G(0) → c1(G(f(0)), F(0))

S tuples:

F(f(z0)) → c(F(z0))
G(0) → c1(G(f(0)), F(0))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c, c1

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

Removed 1 trailing nodes:

G(0) → c1(G(f(0)), F(0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(z0)
g(0) → g(f(0))

Tuples:

F(f(z0)) → c(F(z0))

S tuples:

F(f(z0)) → c(F(z0))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c

(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.

F(f(z0)) → c(F(z0))

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

F(f(z0)) → c(F(z0))

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

POL(F(x₁)) = x₁   
POL(c(x₁)) = x₁   
POL(f(x₁)) = [1] + [4]x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(z0)) → f(z0)
g(0) → g(f(0))

Tuples:

F(f(z0)) → c(F(z0))

S tuples:none
K tuples:

F(f(z0)) → c(F(z0))

Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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