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

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

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
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:

FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
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:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
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 2 less nodes.

(4) Obligation:

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

ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
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.


FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(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__first(x1, x2)  =  n__first(x1, x2)
FIRST(x1, x2)  =  FIRST(x1, x2)
activate(x1)  =  x1
s(x1)  =  x1
cons(x1, x2)  =  x2
n__from(x1)  =  n__from(x1)
n__s(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
0  =  0
nil  =  nil
from(x1)  =  from(x1)

Lexicographic Path Order [LPO].
Precedence:
[nfirst2, FIRST2, first2] > [0, nil] > [nfrom1, from1]


The following usable rules [FROCOS05] were oriented:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

(6) Obligation:

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

ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
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) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(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:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
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) 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)  =  ACTIVATE(x1)
n__s(x1)  =  n__s(x1)
first(x1, x2)  =  x2
0  =  0
nil  =  nil
s(x1)  =  s(x1)
cons(x1, x2)  =  x1
n__first(x1, x2)  =  x2
activate(x1)  =  activate(x1)
from(x1)  =  x1
n__from(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
activate1 > s1 > [ACTIVATE1, ns1, 0, nil]


The following usable rules [FROCOS05] were oriented:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

(10) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
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.

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) TRUE