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, f(b, f(a, x))) → f(a, f(b, f(b, f(a, x))))
f(b, f(b, f(b, x))) → f(b, f(b, x))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(a, f(b, f(a, z0))) → c(F(a, f(b, f(b, f(a, z0)))), F(b, f(b, f(a, z0))), F(b, f(a, z0)), F(a, z0))
F(b, f(b, f(b, z0))) → c1(F(b, f(b, z0)), F(b, z0))

S tuples:

F(a, f(b, f(a, z0))) → c(F(a, f(b, f(b, f(a, z0)))), F(b, f(b, f(a, z0))), F(b, f(a, z0)), F(a, z0))
F(b, f(b, f(b, z0))) → c1(F(b, f(b, z0)), F(b, z0))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

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

Use narrowing to replace F(a, f(b, f(a, z0))) → c(F(a, f(b, f(b, f(a, z0)))), F(b, f(b, f(a, z0))), F(b, f(a, z0)), F(a, z0)) by

F(a, f(b, f(a, x0))) → c(F(a, x0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(b, f(b, f(b, z0))) → c1(F(b, f(b, z0)), F(b, z0))
F(a, f(b, f(a, x0))) → c(F(a, x0))

S tuples:

F(b, f(b, f(b, z0))) → c1(F(b, f(b, z0)), F(b, z0))
F(a, f(b, f(a, x0))) → c(F(a, x0))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, 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(b, f(b, f(b, z0))) → c1(F(b, f(b, z0)), F(b, z0))
F(a, f(b, f(a, x0))) → c(F(a, x0))

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

F(b, f(b, f(b, z0))) → c1(F(b, f(b, z0)), F(b, z0))
F(a, f(b, f(a, x0))) → c(F(a, x0))

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

POL(F(x₁, x₂)) = x₁ + [4]x₂   
POL(a) = 0   
POL(b) = [4]   
POL(c(x₁)) = x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(f(x₁, x₂)) = [2]x₁ + [4]x₂   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(b, f(b, f(b, z0))) → c1(F(b, f(b, z0)), F(b, z0))
F(a, f(b, f(a, x0))) → c(F(a, x0))

S tuples:none
K tuples:

F(b, f(b, f(b, z0))) → c1(F(b, f(b, z0)), F(b, z0))
F(a, f(b, f(a, x0))) → c(F(a, x0))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1, c

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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