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:

a → g(c)
g(a) → b
f(g(X), b) → f(a, X)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

a → g(c)
g(a) → b
f(g(z0), b) → f(a, z0)

Tuples:

A → c1(G(c))
F(g(z0), b) → c3(F(a, z0), A)

S tuples:

A → c1(G(c))
F(g(z0), b) → c3(F(a, z0), A)

K tuples:none
Defined Rule Symbols:

a, g, f

Defined Pair Symbols:

A, F

Compound Symbols:

c1, c3

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

Removed 1 trailing nodes:

A → c1(G(c))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a → g(c)
g(a) → b
f(g(z0), b) → f(a, z0)

Tuples:

F(g(z0), b) → c3(F(a, z0), A)

S tuples:

F(g(z0), b) → c3(F(a, z0), A)

K tuples:none
Defined Rule Symbols:

a, g, f

Defined Pair Symbols:

F

Compound Symbols:

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.

F(g(z0), b) → c3(F(a, z0), A)

We considered the (Usable) Rules:

a → g(c)

And the Tuples:

F(g(z0), b) → c3(F(a, z0), A)

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

POL(A) = 0   
POL(F(x₁, x₂)) = x₁ + [2]x₂   
POL(a) = 0   
POL(b) = [1]   
POL(c) = 0   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(g(x₁)) = [2]x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

a → g(c)
g(a) → b
f(g(z0), b) → f(a, z0)

Tuples:

F(g(z0), b) → c3(F(a, z0), A)

S tuples:none
K tuples:

F(g(z0), b) → c3(F(a, z0), A)

Defined Rule Symbols:

a, g, f

Defined Pair Symbols:

F

Compound Symbols:

c3

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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