Termination w.r.t. Q proof of TRS/nontermin/CSREx2

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

  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

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

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

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

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

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

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

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

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.

LEN(cons(X, Z)) → LEN(Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LEN(x1) = LEN(x1)
cons(x1, x2) = cons(x1, x2)

Recursive Path Order [2].
Precedence:

cons₂ > LEN₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

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.

ADD(s(X), Y) → ADD(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2) = x1
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

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.

FST(s(X), cons(Y, Z)) → FST(X, Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
FST(x1, x2) = FST(x2)
s(x1) = s
cons(x1, x2) = cons(x1, x2)

Recursive Path Order [2].
Precedence:

s > FST₁
cons₂ > FST₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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))

The set Q consists of the following terms:

fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))

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