The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(2^n, INF).

(0) Obligation:

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

f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(activate(X))
activate(X) → X

Rewrite Strategy: FULL

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

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

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

f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(activate(X))
activate(X) → X

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
activate(n__f(n__g(X26_0), X2)) →⁺ f(activate(X26_0), n__f(n__g(activate(X26_0)), activate(X2)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [X26_0 / n__f(n__g(X26_0), X2)].
The result substitution is [ ].

The rewrite sequence
activate(n__f(n__g(X26_0), X2)) →⁺ f(activate(X26_0), n__f(n__g(activate(X26_0)), activate(X2)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0,0].
The pumping substitution is [X26_0 / n__f(n__g(X26_0), X2)].
The result substitution is [ ].

(4) BOUNDS(2^n, INF)