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(a, y) → f(y, g(y))
g(a) → b
g(b) → b

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(a, z0) → c(F(z0, g(z0)), G(z0))

S tuples:

F(a, z0) → c(F(z0, g(z0)), G(z0))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(a, z0) → c(F(z0, g(z0)))

S tuples:

F(a, z0) → c(F(z0, g(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(a, z0) → c(F(z0, g(z0)))

We considered the (Usable) Rules:

g(a) → b
g(b) → b

And the Tuples:

F(a, z0) → c(F(z0, g(z0)))

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

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

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(a, z0) → c(F(z0, g(z0)))

S tuples:none
K tuples:

F(a, z0) → c(F(z0, g(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))