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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 31 ms)
↳8 CdtProblem
↳9 SIsEmptyProof (⇔, 0 ms)
↳10 BOUNDS(O(1), O(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:

ACTIVATE(n__a) → c5(A)
F(n__f(n__a)) → c(F(n__g(n__f(n__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) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(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(ACTIVATE(z0))
S tuples:

ACTIVATE(n__g(z0)) → c6(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

f, a, g, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6

(7) 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(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:

ACTIVATE(n__g(z0)) → c6(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVATE(x1)) = [5]x1   
POL(c6(x1)) = x1   
POL(n__g(x1)) = [1] + x1   

(8) 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(ACTIVATE(z0))
S tuples:none
K tuples:

ACTIVATE(n__g(z0)) → c6(ACTIVATE(z0))
Defined Rule Symbols:

f, a, g, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))