Termination w.r.t. Q proof of /tmp/tmpjP9pE

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:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

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:

DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))
REVERSE(xs) → REV(xs, nil)
LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))
REV(xs, ys) → ISEMPTY(xs)
REV(xs, ys) → LAST(xs)
IF(false, xs, ys, zs) → REV(xs, ys)
REV(xs, ys) → DROPLAST(xs)
REV(xs, ys) → APPEND(ys, last(xs))
REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
APPEND(cons(x, xs), ys) → APPEND(xs, ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

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:

DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))
REVERSE(xs) → REV(xs, nil)
LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))
REV(xs, ys) → ISEMPTY(xs)
REV(xs, ys) → LAST(xs)
IF(false, xs, ys, zs) → REV(xs, ys)
REV(xs, ys) → DROPLAST(xs)
REV(xs, ys) → APPEND(ys, last(xs))
REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
APPEND(cons(x, xs), ys) → APPEND(xs, ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

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 with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

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.

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

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

POL(cons(x₁, x₂)) = 1/4 + (7/2)x_2
POL(APPEND(x₁, x₂)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
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:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

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:

DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

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.

DROPLAST(cons(x, cons(y, ys))) → DROPLAST(cons(y, ys))

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

POL(cons(x₁, x₂)) = 1/4 + (9/4)x_1 + (2)x_2
POL(DROPLAST(x₁)) = (2)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

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

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

            ↳ QDPOrderProof

          ↳ QDP

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

LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

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.

LAST(cons(x, cons(y, ys))) → LAST(cons(y, ys))

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

POL(cons(x₁, x₂)) = 1/4 + (9/4)x_1 + (2)x_2
POL(LAST(x₁)) = (2)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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

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:

IF(false, xs, ys, zs) → REV(xs, ys)
REV(xs, ys) → IF(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)

The TRS R consists of the following rules:

isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
last(cons(x, nil)) → x
last(cons(x, cons(y, ys))) → last(cons(y, ys))
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
reverse(xs) → rev(xs, nil)
rev(xs, ys) → if(isEmpty(xs), dropLast(xs), append(ys, last(xs)), ys)
if(true, xs, ys, zs) → zs
if(false, xs, ys, zs) → rev(xs, ys)

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