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(a) → g(b)
b → f(a, a)
f(a, a) → g(d)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(a) → c(G(b), B)
B → c1(F(a, a))
F(a, a) → c2(G(d))

S tuples:

G(a) → c(G(b), B)
B → c1(F(a, a))
F(a, a) → c2(G(d))

K tuples:none
Defined Rule Symbols:

g, b, f

Defined Pair Symbols:

G, B, F

Compound Symbols:

c, c1, c2

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

Removed 2 trailing nodes:

F(a, a) → c2(G(d))
B → c1(F(a, a))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(a) → c(G(b), B)

S tuples:

G(a) → c(G(b), B)

K tuples:none
Defined Rule Symbols:

g, b, f

Defined Pair Symbols:

G

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.

G(a) → c(G(b), B)

We considered the (Usable) Rules:

b → f(a, a)
f(a, a) → g(d)

And the Tuples:

G(a) → c(G(b), B)

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

POL(B) = 0   
POL(G(x₁)) = x₁   
POL(a) = [1]   
POL(b) = 0   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(d) = 0   
POL(f(x₁, x₂)) = 0   
POL(g(x₁)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

G(a) → c(G(b), B)

S tuples:none
K tuples:

G(a) → c(G(b), B)

Defined Rule Symbols:

g, b, f

Defined Pair Symbols:

G

Compound Symbols:

c

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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