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

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

(3) 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(a, f(a, a)), z0) → c(F(z0, f(f(a, a), a)), F(f(a, a), a), F(a, a))

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

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

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

POL(F(x₁, x₂)) = [2]x₁ + [2]x₂   
POL(a) = 0   
POL(c(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(f(x₁, x₂)) = [1] + [4]x₂   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:none
K tuples:

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

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(6) BOUNDS(O(1), O(1))