### (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(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0n__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 Cpx (relative) TRS 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(0)) → 0
s(z0) → n__s(z0)
0n__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')
F(z0) → c2
P(s(0)) → c3(0')
S(z0) → c4
0'c5
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')
ACTIVATE(z0) → c9
S tuples:

F(0) → c(0')
F(s(0)) → c1(F(p(s(0))), P(s(0)), S(0), 0')
F(z0) → c2
P(s(0)) → c3(0')
S(z0) → c4
0'c5
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')
ACTIVATE(z0) → c9
K tuples:none
Defined Rule Symbols:

f, p, s, 0, activate

Defined Pair Symbols:

F, P, S, 0', ACTIVATE

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9

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

Removed 8 trailing nodes:

F(0) → c(0')
F(s(0)) → c1(F(p(s(0))), P(s(0)), S(0), 0')
0'c5
F(z0) → c2
P(s(0)) → c3(0')
S(z0) → c4
ACTIVATE(z0) → c9
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(0)) → 0
s(z0) → n__s(z0)
0n__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) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing tuple parts

### (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(0)) → 0
s(z0) → n__s(z0)
0n__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(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:

ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:

f, p, s, 0, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6, c7

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:

ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6, c7

### (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)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:

ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACTIVATE(x1)) = [4]x1
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(n__f(x1)) = [1] + x1
POL(n__s(x1)) = [4] + x1

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:none
K tuples:

ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c6, c7

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

The set S is empty