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

Tuples:

F(n__f(n__a)) → c(F(n__g(n__f(n__a))))
ACTIVATE(n__f(z0)) → c4(F(z0))
ACTIVATE(n__a) → c5(A)
ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))

S tuples:

F(n__f(n__a)) → c(F(n__g(n__f(n__a))))
ACTIVATE(n__f(z0)) → c4(F(z0))
ACTIVATE(n__a) → c5(A)
ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

f, a, g, activate

Defined Pair Symbols:

F, ACTIVATE

Compound Symbols:

c, c4, c5, c6

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

Removed 3 trailing nodes:

F(n__f(n__a)) → c(F(n__g(n__f(n__a))))
ACTIVATE(n__a) → c5(A)
ACTIVATE(n__f(z0)) → c4(F(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))

S tuples:

ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

f, a, g, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6

(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.

ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))

We considered the (Usable) Rules:

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

And the Tuples:

ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))

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

POL(ACTIVATE(x₁)) = [2]x₁   
POL(G(x₁)) = [2]   
POL(a) = [4]   
POL(activate(x₁)) = [1]   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(f(x₁)) = [5] + x₁   
POL(g(x₁)) = [3]   
POL(n__a) = [2]   
POL(n__f(x₁)) = [3]   
POL(n__g(x₁)) = [4] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))

S tuples:none
K tuples:

ACTIVATE(n__g(z0)) → c6(G(activate(z0)), ACTIVATE(z0))

Defined Rule Symbols:

f, a, g, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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