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 CdtUsableRulesProof (⇔, 0 ms)
↳8 CdtProblem
↳9 CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))), 39 ms)
↳10 CdtProblem
↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳12 BOUNDS(O(1), O(1))

(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 Cpx (relative) TRS 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))))
F(z0) → c1
A → c2
ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__a) → c4(A)
ACTIVATE(z0) → c5
S tuples:

F(f(a)) → c(F(g(n__f(n__a))))
F(z0) → c1
A → c2
ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__a) → c4(A)
ACTIVATE(z0) → c5
K tuples:none
Defined Rule Symbols:

f, a, activate

Defined Pair Symbols:

F, A, ACTIVATE

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 5 trailing nodes:

F(f(a)) → c(F(g(n__f(n__a))))
F(z0) → c1
ACTIVATE(z0) → c5
ACTIVATE(n__a) → c4(A)
A → c2

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

Removed 1 trailing tuple parts

(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(ACTIVATE(z0))
S tuples:

ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

f, a, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
S tuples:

ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3

(9) CdtRuleRemovalProof (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__f(z0)) → c3(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:

ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVATE(x1)) = [5]x1   
POL(c3(x1)) = x1   
POL(n__f(x1)) = [1] + x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
S tuples:none
K tuples:

ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c3

(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(12) BOUNDS(O(1), O(1))