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(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(X)) → X
f(X) → n__f(X)
s(X) → n__s(X)
0 → n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(z0)) → z0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0

Tuples:

F(0) → c(0')
F(s(0)) → c1(F(p(s(0))), P(s(0)), S(0), 0')
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__0) → c8(0')

S tuples:

F(0) → c(0')
F(s(0)) → c1(F(p(s(0))), P(s(0)), S(0), 0')
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__0) → c8(0')

K tuples:none
Defined Rule Symbols:

f, p, s, 0, activate

Defined Pair Symbols:

F, ACTIVATE

Compound Symbols:

c, c1, c6, c7, c8

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

Removed 3 trailing nodes:

F(0) → c(0')
F(s(0)) → c1(F(p(s(0))), P(s(0)), S(0), 0')
ACTIVATE(n__0) → c8(0')

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(z0)) → z0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0

Tuples:

ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))

S tuples:

ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

f, p, s, 0, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6, c7

(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__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))

We considered the (Usable) Rules:

activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0
0 → n__0
s(z0) → n__s(z0)
f(z0) → n__f(z0)

And the Tuples:

ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))

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

POL(0) = 0   
POL(ACTIVATE(x₁)) = [2]x₁   
POL(F(x₁)) = 0   
POL(S(x₁)) = 0   
POL(activate(x₁)) = 0   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c7(x₁, x₂)) = x₁ + x₂   
POL(f(x₁)) = 0   
POL(n__0) = 0   
POL(n__f(x₁)) = [1] + x₁   
POL(n__s(x₁)) = [1] + x₁   
POL(s(x₁)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(z0)) → z0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0

Tuples:

ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))

S tuples:none
K tuples:

ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))

Defined Rule Symbols:

f, p, s, 0, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6, c7

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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