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 (ComplexityIfPolyImplication, 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))), 34 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(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
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(s(0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
activate(n__f(z0)) → f(z0)
activate(z0) → z0
Tuples:

F(0) → c
F(s(0)) → c1(F(p(s(0))), P(s(0)))
F(z0) → c2
P(s(0)) → c3
ACTIVATE(n__f(z0)) → c4(F(z0))
ACTIVATE(z0) → c5
S tuples:

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

f, p, activate

Defined Pair Symbols:

F, P, ACTIVATE

Compound Symbols:

c, c1, c2, c3, c4, c5

(3) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

ACTIVATE(n__f(z0)) → c4(F(z0))
Removed 4 trailing nodes:

F(0) → c
F(z0) → c2
P(s(0)) → c3
ACTIVATE(z0) → c5

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
activate(n__f(z0)) → f(z0)
activate(z0) → z0
Tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0)))
S tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0)))
K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

F

Compound Symbols:

c1

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
activate(n__f(z0)) → f(z0)
activate(z0) → z0
Tuples:

F(s(0)) → c1(F(p(s(0))))
S tuples:

F(s(0)) → c1(F(p(s(0))))
K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

F

Compound Symbols:

c1

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(0)) → 0
Tuples:

F(s(0)) → c1(F(p(s(0))))
S tuples:

F(s(0)) → c1(F(p(s(0))))
K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:

F

Compound Symbols:

c1

(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.

F(s(0)) → c1(F(p(s(0))))
We considered the (Usable) Rules:

p(s(0)) → 0
And the Tuples:

F(s(0)) → c1(F(p(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [3]   
POL(F(x1)) = [4]x1   
POL(c1(x1)) = x1   
POL(p(x1)) = [3]   
POL(s(x1)) = [5] + x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(0)) → 0
Tuples:

F(s(0)) → c1(F(p(s(0))))
S tuples:none
K tuples:

F(s(0)) → c1(F(p(s(0))))
Defined Rule Symbols:

p

Defined Pair Symbols:

F

Compound Symbols:

c1

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

The set S is empty

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