Termination w.r.t. Q proof of /tmp/tmp9l4LNl/Ex15_Luc98

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ADD(s(X), Y) → S(n__add(activate(X), activate(Y)))
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__s(X)) → S(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
IF(true, X, Y) → ACTIVATE(X)
FROM(X) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
AND(true, X) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → FROM(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(0, X) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

ADD(s(X), Y) → S(n__add(activate(X), activate(Y)))
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__s(X)) → S(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
IF(true, X, Y) → ACTIVATE(X)
FROM(X) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
AND(true, X) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → FROM(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(0, X) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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(X)
FROM(X) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → FROM(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(0, X) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

ADD(0, X) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(n__first(x₁, x₂)) = (4)x_1 + (7/4)x_2
POL(from(x₁)) = (4)x_1
POL(activate(x₁)) = x_1
POL(n__s(x₁)) = (3/4)x_1
POL(0) = 1/4
POL(first(x₁, x₂)) = (4)x_1 + (7/4)x_2
POL(FROM(x₁)) = (5/2)x_1
POL(add(x₁, x₂)) = (3)x_1 + x_2
POL(FIRST(x₁, x₂)) = (2)x_1 + (2)x_2
POL(cons(x₁, x₂)) = (3/4)x_1 + (3/4)x_2
POL(n__add(x₁, x₂)) = (3)x_1 + x_2
POL(n__from(x₁)) = (4)x_1
POL(s(x₁)) = (3/4)x_1
POL(ADD(x₁, x₂)) = (4)x_1 + (3/2)x_2
POL(ACTIVATE(x₁)) = (3/2)x_1
POL(nil) = 1

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), X2)
ADD(s(X), Y) → ACTIVATE(Y)
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(n__first(x₁, x₂)) = x_1 + (2)x_2
POL(from(x₁)) = x_1
POL(activate(x₁)) = x_1
POL(n__s(x₁)) = x_1
POL(0) = 4
POL(first(x₁, x₂)) = x_1 + (2)x_2
POL(FROM(x₁)) = (1/4)x_1
POL(add(x₁, x₂)) = 9/4 + (4)x_1 + (3/2)x_2
POL(FIRST(x₁, x₂)) = (1/4)x_1 + (1/2)x_2
POL(cons(x₁, x₂)) = (1/2)x_1 + (1/2)x_2
POL(n__add(x₁, x₂)) = 9/4 + (4)x_1 + (3/2)x_2
POL(n__from(x₁)) = x_1
POL(s(x₁)) = x_1
POL(ADD(x₁, x₂)) = 1/4 + (1/4)x_1 + (1/4)x_2
POL(ACTIVATE(x₁)) = (1/4)x_1
POL(nil) = 15/4

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__from(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(n__first(x₁, x₂)) = (4)x_1 + (4)x_2
POL(from(x₁)) = 1/2 + (2)x_1
POL(activate(x₁)) = x_1
POL(n__s(x₁)) = (1/2)x_1
POL(0) = 0
POL(first(x₁, x₂)) = (4)x_1 + (4)x_2
POL(FROM(x₁)) = (5/4)x_1
POL(FIRST(x₁, x₂)) = (4)x_1 + (4)x_2
POL(cons(x₁, x₂)) = (1/4)x_1 + (1/2)x_2
POL(add(x₁, x₂)) = 1/2 + x_1 + (5/2)x_2
POL(n__add(x₁, x₂)) = 1/2 + x_1 + (5/2)x_2
POL(n__from(x₁)) = 1/2 + (2)x_1
POL(s(x₁)) = (1/2)x_1
POL(ACTIVATE(x₁)) = x_1
POL(nil) = 0

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)

Used ordering: Polynomial interpretation [25,35]:

POL(n__first(x₁, x₂)) = (3/2)x_1 + (2)x_2
POL(from(x₁)) = 2 + x_1
POL(activate(x₁)) = x_1
POL(n__s(x₁)) = (3/4)x_1
POL(first(x₁, x₂)) = (3/2)x_1 + (2)x_2
POL(0) = 0
POL(FIRST(x₁, x₂)) = 1/4 + (3/4)x_1 + x_2
POL(cons(x₁, x₂)) = 1/2 + (1/2)x_1 + (1/2)x_2
POL(add(x₁, x₂)) = (5/4)x_2
POL(n__add(x₁, x₂)) = (5/4)x_2
POL(n__from(x₁)) = 2 + x_1
POL(s(x₁)) = (3/4)x_1
POL(ACTIVATE(x₁)) = 1/2 + (1/2)x_1
POL(nil) = 0

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ UsableRulesProof

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

ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QDPSizeChangeProof

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

ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
The graph contains the following edges 1 > 1
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
The graph contains the following edges 1 > 1