(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
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:
2ND(cons(X, X1)) → 2ND(cons1(X, activate(X1)))
2ND(cons(X, X1)) → ACTIVATE(X1)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
The TRS R consists of the following rules:
2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
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 1 SCC with 4 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
The TRS R consists of the following rules:
2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVATE(
x0,
x1) =
ACTIVATE(
x1)
Tags:
ACTIVATE has argument tags [1,0] and root tag 0
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
ACTIVATE(
x1) =
ACTIVATE
n__s(
x1) =
n__s(
x1)
n__from(
x1) =
x1
Recursive path order with status [RPO].
Quasi-Precedence:
trivial
Status:
ACTIVATE: multiset
ns1: multiset
The following usable rules [FROCOS05] were oriented:
none
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__from(X)) → ACTIVATE(X)
The TRS R consists of the following rules:
2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
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__from(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVATE(
x0,
x1) =
ACTIVATE(
x1)
Tags:
ACTIVATE has argument tags [1,0] and root tag 0
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
ACTIVATE(
x1) =
ACTIVATE
n__from(
x1) =
n__from(
x1)
Recursive path order with status [RPO].
Quasi-Precedence:
[ACTIVATE, nfrom1]
Status:
ACTIVATE: multiset
nfrom1: multiset
The following usable rules [FROCOS05] were oriented:
none
(8) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(10) TRUE