The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

f(f(a)) → f(g(n__f(n__a)))
f(X) → n__f(X)
a → n__a
activate(n__f(X)) → f(activate(X))
activate(n__a) → a
activate(X) → X

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(g(n__f(n__a)))
f(z0) → n__f(z0)
a → n__a
activate(n__f(z0)) → f(activate(z0))
activate(n__a) → a
activate(z0) → z0

Tuples:

F(f(a)) → c(F(g(n__f(n__a))))
ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__a) → c4(A)

S tuples:

F(f(a)) → c(F(g(n__f(n__a))))
ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__a) → c4(A)

K tuples:none
Defined Rule Symbols:

f, a, activate

Defined Pair Symbols:

F, ACTIVATE

Compound Symbols:

c, c3, c4

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

Removed 2 trailing nodes:

F(f(a)) → c(F(g(n__f(n__a))))
ACTIVATE(n__a) → c4(A)

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(a)) → f(g(n__f(n__a)))
f(z0) → n__f(z0)
a → n__a
activate(n__f(z0)) → f(activate(z0))
activate(n__a) → a
activate(z0) → z0

Tuples:

ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))

S tuples:

ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

f, a, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))

We considered the (Usable) Rules:

activate(n__f(z0)) → f(activate(z0))
activate(n__a) → a
activate(z0) → z0
a → n__a
f(z0) → n__f(z0)

And the Tuples:

ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))

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

POL(ACTIVATE(x₁)) = x₁²   
POL(F(x₁)) = 0   
POL(a) = 0   
POL(activate(x₁)) = 0   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(f(x₁)) = 0   
POL(n__a) = 0   
POL(n__f(x₁)) = [2] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(f(a)) → f(g(n__f(n__a)))
f(z0) → n__f(z0)
a → n__a
activate(n__f(z0)) → f(activate(z0))
activate(n__a) → a
activate(z0) → z0

Tuples:

ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))

S tuples:none
K tuples:

ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))

Defined Rule Symbols:

f, a, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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