Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

Q restricted rewrite system:
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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(s(0)) → F(p(s(0)))
F(s(0)) → P(s(0))
ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__0) → 0¹

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(s(0)) → F(p(s(0)))

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__f(X)) → ACTIVATE(X)

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__f(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE(x1)
n__s(x1)  =  x1
n__f(x1)  =  n__f(x1)

Lexicographic Path Order [LPO].
Precedence:

n_f1 > ACTIVATE₁

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__s(X)) → ACTIVATE(X)

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__s(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__s(x1)  =  n__s(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

Q DP problem:
P is empty.
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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(12) TRUE