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(f(a, x), a)) → f(f(a, f(a, x)), a)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:

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

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

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

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

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

POL(F(x₁, x₂)) = [2] + [5]x₂   
POL(a) = [4]   
POL(c(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(f(x₁, x₂)) = [5] + [4]x₁ + [5]x₂   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:none
K tuples:

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

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))