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:

a(a(x)) → b(b(x))
b(b(a(x))) → a(b(b(x)))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

a(a(z0)) → b(b(z0))
b(b(a(z0))) → a(b(b(z0)))

Tuples:

A(a(z0)) → c(B(b(z0)), B(z0))
B(b(a(z0))) → c1(A(b(b(z0))), B(b(z0)), B(z0))

S tuples:

A(a(z0)) → c(B(b(z0)), B(z0))
B(b(a(z0))) → c1(A(b(b(z0))), B(b(z0)), B(z0))

K tuples:none
Defined Rule Symbols:

a, b

Defined Pair Symbols:

A, B

Compound Symbols:

c, c1

(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.

A(a(z0)) → c(B(b(z0)), B(z0))
B(b(a(z0))) → c1(A(b(b(z0))), B(b(z0)), B(z0))

We considered the (Usable) Rules:

b(b(a(z0))) → a(b(b(z0)))
a(a(z0)) → b(b(z0))

And the Tuples:

A(a(z0)) → c(B(b(z0)), B(z0))
B(b(a(z0))) → c1(A(b(b(z0))), B(b(z0)), B(z0))

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

POL(A(x₁)) = x₁   
POL(B(x₁)) = [2]x₁   
POL(a(x₁)) = [1] + [5]x₁   
POL(b(x₁)) = x₁   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂, x₃)) = x₁ + x₂ + x₃   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

a(a(z0)) → b(b(z0))
b(b(a(z0))) → a(b(b(z0)))

Tuples:

A(a(z0)) → c(B(b(z0)), B(z0))
B(b(a(z0))) → c1(A(b(b(z0))), B(b(z0)), B(z0))

S tuples:none
K tuples:

A(a(z0)) → c(B(b(z0)), B(z0))
B(b(a(z0))) → c1(A(b(b(z0))), B(b(z0)), B(z0))

Defined Rule Symbols:

a, b

Defined Pair Symbols:

A, B

Compound Symbols:

c, c1

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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