Termination w.r.t. Q proof of /tmp/tmpxb2ZKl/Ex2

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

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:

FST(s(X), cons(Y, Z)) → FST(X, Z)
LEN(cons(X, Z)) → LEN(Z)
ADD(s(X), Y) → ADD(X, Y)
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

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:

FST(s(X), cons(Y, Z)) → FST(X, Z)
LEN(cons(X, Z)) → LEN(Z)
ADD(s(X), Y) → ADD(X, Y)
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LEN(cons(X, Z)) → LEN(Z)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

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.

LEN(cons(X, Z)) → LEN(Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 1/4 + (4)x_2
POL(LEN(x₁)) = (4)x_1

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

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(s(X), Y) → ADD(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x₁)) = 1/4 + (2)x_1
POL(ADD(x₁, x₂)) = (1/4)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

FST(s(X), cons(Y, Z)) → FST(X, Z)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

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.

FST(s(X), cons(Y, Z)) → FST(X, Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 4 + (2)x_1 + (4)x_2
POL(FST(x₁, x₂)) = (4)x_2
POL(s(x₁)) = 0

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))

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.