Termination w.r.t. Q proof of TRS/TRCSREx15_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:

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

AND₂(true, X) -> ACTIVATE₁(X)
ADD₂(0, X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> S₁(X)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
IF₃(true, X, Y) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> S₁(n__add₂(activate₁(X), activate₁(Y)))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Y)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
ADD₂(s₁(X), Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
FROM₁(X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)

The TRS R consists of the following rules:

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(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:

AND₂(true, X) -> ACTIVATE₁(X)
ADD₂(0, X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> S₁(X)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
IF₃(true, X, Y) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> S₁(n__add₂(activate₁(X), activate₁(Y)))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Y)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
ADD₂(s₁(X), Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
FROM₁(X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)

The TRS R consists of the following rules:

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

ADD₂(0, X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
FROM₁(X) -> ACTIVATE₁(X)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Y)

The TRS R consists of the following rules:

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ADD₂(0, X) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
FROM₁(X) -> ACTIVATE₁(X)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Y)

The remaining pairs can at least be oriented weakly.

ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)

Used ordering: Polynomial interpretation [21]:

POL(0) = 1
POL(ACTIVATE₁(x₁)) = x₁
POL(ADD₂(x₁, x₂)) = 2·x₁ + x₂
POL(FIRST₂(x₁, x₂)) = 1 + x₁ + x₂
POL(FROM₁(x₁)) = 1 + 2·x₁
POL(cons₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(n__add₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(n__first₂(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(n__from₁(x₁)) = 2 + 2·x₁
POL(s₁(x₁)) = 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
ADD₂(s₁(X), Y) -> ACTIVATE₁(Y)

The TRS R consists of the following rules:

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
ADD₂(s₁(X), Y) -> ACTIVATE₁(Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = 3 + 3·x₁
POL(ADD₂(x₁, x₂)) = 3 + 3·x₁ + 3·x₂
POL(n__add₂(x₁, x₂)) = 3 + x₁ + 2·x₂
POL(s₁(x₁)) = 3 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

and₂(true, X) -> activate₁(X)
and₂(false, Y) -> false
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
add₂(0, X) -> activate₁(X)
add₂(s₁(X), Y) -> s₁(n__add₂(activate₁(X), activate₁(Y)))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(activate₁(Y), n__first₂(activate₁(X), activate₁(Z)))
from₁(X) -> cons₂(activate₁(X), n__from₁(n__s₁(activate₁(X))))
add₂(X1, X2) -> n__add₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__add₂(X1, X2)) -> add₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__s₁(X)) -> s₁(X)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.