Termination w.r.t. Q proof of TRS/secret06double.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:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

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:

PERMUTE(x, y, a) → ISZERO(x)
PLUS(x, s(s(y))) → PLUS(s(x), y)
PLUS(s(x), y) → PLUS(x, s(y))
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
ACK(s(x), 0) → ACK(x, s(0))
ACK(0, x) → PLUS(x, s(0))
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)
ACK(s(x), s(y)) → ACK(s(x), y)
DOUBLE(x) → PERMUTE(x, x, a)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
PERMUTE(false, x, b) → ACK(x, x)
PERMUTE(y, x, c) → PERMUTE(x, y, a)
PERMUTE(false, x, b) → P(x)

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

PERMUTE(x, y, a) → ISZERO(x)
PLUS(x, s(s(y))) → PLUS(s(x), y)
PLUS(s(x), y) → PLUS(x, s(y))
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
ACK(s(x), 0) → ACK(x, s(0))
ACK(0, x) → PLUS(x, s(0))
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)
ACK(s(x), s(y)) → ACK(s(x), y)
DOUBLE(x) → PERMUTE(x, x, a)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
PERMUTE(false, x, b) → ACK(x, x)
PERMUTE(y, x, c) → PERMUTE(x, y, a)
PERMUTE(false, x, b) → P(x)

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

PERMUTE(x, y, a) → ISZERO(x)
PLUS(x, s(s(y))) → PLUS(s(x), y)
PLUS(s(x), y) → PLUS(x, s(y))
ACK(s(x), 0) → ACK(x, s(0))
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
ACK(0, x) → PLUS(x, s(0))
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)
ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
DOUBLE(x) → PERMUTE(x, x, a)
PERMUTE(false, x, b) → ACK(x, x)
PERMUTE(y, x, c) → PERMUTE(x, y, a)
PERMUTE(false, x, b) → P(x)

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(x, s(s(y))) → PLUS(s(x), y)
PLUS(s(x), y) → PLUS(x, s(y))

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
ACK(s(x), 0) → ACK(x, s(0))

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

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.

ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
ACK(s(x), 0) → ACK(x, s(0))

The remaining pairs can at least be oriented weakly.

ACK(s(x), s(y)) → ACK(s(x), y)

Used ordering: Combined order from the following AFS and order.
ACK(x1, x2) = x1
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

ACK(s(x), s(y)) → ACK(s(x), y)

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

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.

ACK(s(x), s(y)) → ACK(s(x), y)

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

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

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PERMUTE(y, x, c) → PERMUTE(x, y, a)
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)

The TRS R consists of the following rules:

double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false

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