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(X) → g(n__h(n__f(X)))
h(X) → n__h(X)
f(X) → n__f(X)
activate(n__h(X)) → h(activate(X))
activate(n__f(X)) → f(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(z0) → g(n__h(n__f(z0)))
f(z0) → n__f(z0)
h(z0) → n__h(z0)
activate(n__h(z0)) → h(activate(z0))
activate(n__f(z0)) → f(activate(z0))
activate(z0) → z0

Tuples:

ACTIVATE(n__h(z0)) → c3(H(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(F(activate(z0)), ACTIVATE(z0))

S tuples:

ACTIVATE(n__h(z0)) → c3(H(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(F(activate(z0)), ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

f, h, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3, c4

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

Removed 2 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → g(n__h(n__f(z0)))
f(z0) → n__f(z0)
h(z0) → n__h(z0)
activate(n__h(z0)) → h(activate(z0))
activate(n__f(z0)) → f(activate(z0))
activate(z0) → z0

Tuples:

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))

S tuples:

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

f, h, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3, c4

(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__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))

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

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))

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

POL(ACTIVATE(x₁)) = [4]x₁   
POL(c3(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(n__f(x₁)) = [4] + x₁   
POL(n__h(x₁)) = [1] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(z0) → g(n__h(n__f(z0)))
f(z0) → n__f(z0)
h(z0) → n__h(z0)
activate(n__h(z0)) → h(activate(z0))
activate(n__f(z0)) → f(activate(z0))
activate(z0) → z0

Tuples:

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))

S tuples:none
K tuples:

ACTIVATE(n__h(z0)) → c3(ACTIVATE(z0))
ACTIVATE(n__f(z0)) → c4(ACTIVATE(z0))

Defined Rule Symbols:

f, h, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3, c4

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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