Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE

(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(0)) → 0
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(0)) → 0
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(0)) → 0
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(s(0)) → F(p(s(0)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  F(x1)
s(x1)  =  s(x1)
0  =  0
p(x1)  =  p
f(x1)  =  f(x1)
cons(x1, x2)  =  cons(x1, x2)
n__f(x1)  =  n__f(x1)
n__s(x1)  =  n__s(x1)
n__0  =  n__0
activate(x1)  =  activate(x1)

Recursive Path Order [RPO].
Precedence:
F1 > s1 > p > 0 > n0
F1 > s1 > ns1
activate1 > f1 > s1 > p > 0 > n0
activate1 > f1 > s1 > ns1
activate1 > f1 > cons2
activate1 > f1 > nf1

The following usable rules [FROCOS05] were oriented:

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

(7) 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(0)) → 0
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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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(0)) → 0
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) 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)
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)  =  n__s(x1)
n__f(x1)  =  n__f(x1)
f(x1)  =  f(x1)
0  =  0
cons(x1, x2)  =  cons(x1, x2)
n__0  =  n__0
s(x1)  =  s(x1)
p(x1)  =  x1
activate(x1)  =  activate(x1)

Recursive Path Order [RPO].
Precedence:
activate1 > 0 > s1 > ns1 > ACTIVATE1
activate1 > 0 > s1 > f1 > nf1 > ACTIVATE1
activate1 > 0 > s1 > f1 > cons2
activate1 > 0 > s1 > f1 > n0

The following usable rules [FROCOS05] were oriented:

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

(12) 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(0)) → 0
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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE