Termination w.r.t. Q proof of TRS/secret06reverse.trs

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:

isEmpty₁(nil) -> true
isEmpty₁(cons₂(x, xs)) -> false
last₁(cons₂(x, nil)) -> x
last₁(cons₂(x, cons₂(y, ys))) -> last₁(cons₂(y, ys))
dropLast₁(nil) -> nil
dropLast₁(cons₂(x, nil)) -> nil
dropLast₁(cons₂(x, cons₂(y, ys))) -> cons₂(x, dropLast₁(cons₂(y, ys)))
append₂(nil, ys) -> ys
append₂(cons₂(x, xs), ys) -> cons₂(x, append₂(xs, ys))
reverse₁(xs) -> rev₂(xs, nil)
rev₂(xs, ys) -> if₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
if₄(true, xs, ys, zs) -> zs
if₄(false, xs, ys, zs) -> rev₂(xs, ys)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

isEmpty₁(nil) -> true
isEmpty₁(cons₂(x, xs)) -> false
last₁(cons₂(x, nil)) -> x
last₁(cons₂(x, cons₂(y, ys))) -> last₁(cons₂(y, ys))
dropLast₁(nil) -> nil
dropLast₁(cons₂(x, nil)) -> nil
dropLast₁(cons₂(x, cons₂(y, ys))) -> cons₂(x, dropLast₁(cons₂(y, ys)))
append₂(nil, ys) -> ys
append₂(cons₂(x, xs), ys) -> cons₂(x, append₂(xs, ys))
reverse₁(xs) -> rev₂(xs, nil)
rev₂(xs, ys) -> if₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
if₄(true, xs, ys, zs) -> zs
if₄(false, xs, ys, zs) -> rev₂(xs, ys)

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:

REV₂(xs, ys) -> DROPLAST₁(xs)
REV₂(xs, ys) -> ISEMPTY₁(xs)
REV₂(xs, ys) -> LAST₁(xs)
REV₂(xs, ys) -> IF₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
DROPLAST₁(cons₂(x, cons₂(y, ys))) -> DROPLAST₁(cons₂(y, ys))
REVERSE₁(xs) -> REV₂(xs, nil)
IF₄(false, xs, ys, zs) -> REV₂(xs, ys)
APPEND₂(cons₂(x, xs), ys) -> APPEND₂(xs, ys)
REV₂(xs, ys) -> APPEND₂(ys, last₁(xs))
LAST₁(cons₂(x, cons₂(y, ys))) -> LAST₁(cons₂(y, ys))

The TRS R consists of the following rules:

isEmpty₁(nil) -> true
isEmpty₁(cons₂(x, xs)) -> false
last₁(cons₂(x, nil)) -> x
last₁(cons₂(x, cons₂(y, ys))) -> last₁(cons₂(y, ys))
dropLast₁(nil) -> nil
dropLast₁(cons₂(x, nil)) -> nil
dropLast₁(cons₂(x, cons₂(y, ys))) -> cons₂(x, dropLast₁(cons₂(y, ys)))
append₂(nil, ys) -> ys
append₂(cons₂(x, xs), ys) -> cons₂(x, append₂(xs, ys))
reverse₁(xs) -> rev₂(xs, nil)
rev₂(xs, ys) -> if₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
if₄(true, xs, ys, zs) -> zs
if₄(false, xs, ys, zs) -> rev₂(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:

REV₂(xs, ys) -> DROPLAST₁(xs)
REV₂(xs, ys) -> ISEMPTY₁(xs)
REV₂(xs, ys) -> LAST₁(xs)
REV₂(xs, ys) -> IF₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
DROPLAST₁(cons₂(x, cons₂(y, ys))) -> DROPLAST₁(cons₂(y, ys))
REVERSE₁(xs) -> REV₂(xs, nil)
IF₄(false, xs, ys, zs) -> REV₂(xs, ys)
APPEND₂(cons₂(x, xs), ys) -> APPEND₂(xs, ys)
REV₂(xs, ys) -> APPEND₂(ys, last₁(xs))
LAST₁(cons₂(x, cons₂(y, ys))) -> LAST₁(cons₂(y, ys))

The TRS R consists of the following rules:

isEmpty₁(nil) -> true
isEmpty₁(cons₂(x, xs)) -> false
last₁(cons₂(x, nil)) -> x
last₁(cons₂(x, cons₂(y, ys))) -> last₁(cons₂(y, ys))
dropLast₁(nil) -> nil
dropLast₁(cons₂(x, nil)) -> nil
dropLast₁(cons₂(x, cons₂(y, ys))) -> cons₂(x, dropLast₁(cons₂(y, ys)))
append₂(nil, ys) -> ys
append₂(cons₂(x, xs), ys) -> cons₂(x, append₂(xs, ys))
reverse₁(xs) -> rev₂(xs, nil)
rev₂(xs, ys) -> if₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
if₄(true, xs, ys, zs) -> zs
if₄(false, xs, ys, zs) -> rev₂(xs, ys)

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

APPEND₂(cons₂(x, xs), ys) -> APPEND₂(xs, ys)

The TRS R consists of the following rules:

isEmpty₁(nil) -> true
isEmpty₁(cons₂(x, xs)) -> false
last₁(cons₂(x, nil)) -> x
last₁(cons₂(x, cons₂(y, ys))) -> last₁(cons₂(y, ys))
dropLast₁(nil) -> nil
dropLast₁(cons₂(x, nil)) -> nil
dropLast₁(cons₂(x, cons₂(y, ys))) -> cons₂(x, dropLast₁(cons₂(y, ys)))
append₂(nil, ys) -> ys
append₂(cons₂(x, xs), ys) -> cons₂(x, append₂(xs, ys))
reverse₁(xs) -> rev₂(xs, nil)
rev₂(xs, ys) -> if₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
if₄(true, xs, ys, zs) -> zs
if₄(false, xs, ys, zs) -> rev₂(xs, ys)

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.

APPEND₂(cons₂(x, xs), ys) -> APPEND₂(xs, ys)

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

POL(APPEND₂(x₁, x₂)) = x₁
POL(cons₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] 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:

isEmpty₁(nil) -> true
isEmpty₁(cons₂(x, xs)) -> false
last₁(cons₂(x, nil)) -> x
last₁(cons₂(x, cons₂(y, ys))) -> last₁(cons₂(y, ys))
dropLast₁(nil) -> nil
dropLast₁(cons₂(x, nil)) -> nil
dropLast₁(cons₂(x, cons₂(y, ys))) -> cons₂(x, dropLast₁(cons₂(y, ys)))
append₂(nil, ys) -> ys
append₂(cons₂(x, xs), ys) -> cons₂(x, append₂(xs, ys))
reverse₁(xs) -> rev₂(xs, nil)
rev₂(xs, ys) -> if₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
if₄(true, xs, ys, zs) -> zs
if₄(false, xs, ys, zs) -> rev₂(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:

DROPLAST₁(cons₂(x, cons₂(y, ys))) -> DROPLAST₁(cons₂(y, ys))

The TRS R consists of the following rules:

isEmpty₁(nil) -> true
isEmpty₁(cons₂(x, xs)) -> false
last₁(cons₂(x, nil)) -> x
last₁(cons₂(x, cons₂(y, ys))) -> last₁(cons₂(y, ys))
dropLast₁(nil) -> nil
dropLast₁(cons₂(x, nil)) -> nil
dropLast₁(cons₂(x, cons₂(y, ys))) -> cons₂(x, dropLast₁(cons₂(y, ys)))
append₂(nil, ys) -> ys
append₂(cons₂(x, xs), ys) -> cons₂(x, append₂(xs, ys))
reverse₁(xs) -> rev₂(xs, nil)
rev₂(xs, ys) -> if₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
if₄(true, xs, ys, zs) -> zs
if₄(false, xs, ys, zs) -> rev₂(xs, ys)

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.

DROPLAST₁(cons₂(x, cons₂(y, ys))) -> DROPLAST₁(cons₂(y, ys))

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

POL(DROPLAST₁(x₁)) = 1 + x₁
POL(cons₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] 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:

isEmpty₁(nil) -> true
isEmpty₁(cons₂(x, xs)) -> false
last₁(cons₂(x, nil)) -> x
last₁(cons₂(x, cons₂(y, ys))) -> last₁(cons₂(y, ys))
dropLast₁(nil) -> nil
dropLast₁(cons₂(x, nil)) -> nil
dropLast₁(cons₂(x, cons₂(y, ys))) -> cons₂(x, dropLast₁(cons₂(y, ys)))
append₂(nil, ys) -> ys
append₂(cons₂(x, xs), ys) -> cons₂(x, append₂(xs, ys))
reverse₁(xs) -> rev₂(xs, nil)
rev₂(xs, ys) -> if₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
if₄(true, xs, ys, zs) -> zs
if₄(false, xs, ys, zs) -> rev₂(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:

LAST₁(cons₂(x, cons₂(y, ys))) -> LAST₁(cons₂(y, ys))

The TRS R consists of the following rules:

isEmpty₁(nil) -> true
isEmpty₁(cons₂(x, xs)) -> false
last₁(cons₂(x, nil)) -> x
last₁(cons₂(x, cons₂(y, ys))) -> last₁(cons₂(y, ys))
dropLast₁(nil) -> nil
dropLast₁(cons₂(x, nil)) -> nil
dropLast₁(cons₂(x, cons₂(y, ys))) -> cons₂(x, dropLast₁(cons₂(y, ys)))
append₂(nil, ys) -> ys
append₂(cons₂(x, xs), ys) -> cons₂(x, append₂(xs, ys))
reverse₁(xs) -> rev₂(xs, nil)
rev₂(xs, ys) -> if₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
if₄(true, xs, ys, zs) -> zs
if₄(false, xs, ys, zs) -> rev₂(xs, ys)

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.

LAST₁(cons₂(x, cons₂(y, ys))) -> LAST₁(cons₂(y, ys))

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

POL(LAST₁(x₁)) = 1 + x₁
POL(cons₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] 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:

isEmpty₁(nil) -> true
isEmpty₁(cons₂(x, xs)) -> false
last₁(cons₂(x, nil)) -> x
last₁(cons₂(x, cons₂(y, ys))) -> last₁(cons₂(y, ys))
dropLast₁(nil) -> nil
dropLast₁(cons₂(x, nil)) -> nil
dropLast₁(cons₂(x, cons₂(y, ys))) -> cons₂(x, dropLast₁(cons₂(y, ys)))
append₂(nil, ys) -> ys
append₂(cons₂(x, xs), ys) -> cons₂(x, append₂(xs, ys))
reverse₁(xs) -> rev₂(xs, nil)
rev₂(xs, ys) -> if₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
if₄(true, xs, ys, zs) -> zs
if₄(false, xs, ys, zs) -> rev₂(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:

REV₂(xs, ys) -> IF₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
IF₄(false, xs, ys, zs) -> REV₂(xs, ys)

The TRS R consists of the following rules:

isEmpty₁(nil) -> true
isEmpty₁(cons₂(x, xs)) -> false
last₁(cons₂(x, nil)) -> x
last₁(cons₂(x, cons₂(y, ys))) -> last₁(cons₂(y, ys))
dropLast₁(nil) -> nil
dropLast₁(cons₂(x, nil)) -> nil
dropLast₁(cons₂(x, cons₂(y, ys))) -> cons₂(x, dropLast₁(cons₂(y, ys)))
append₂(nil, ys) -> ys
append₂(cons₂(x, xs), ys) -> cons₂(x, append₂(xs, ys))
reverse₁(xs) -> rev₂(xs, nil)
rev₂(xs, ys) -> if₄(isEmpty₁(xs), dropLast₁(xs), append₂(ys, last₁(xs)), ys)
if₄(true, xs, ys, zs) -> zs
if₄(false, xs, ys, zs) -> rev₂(xs, ys)

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