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:

double₁(x) -> permute₃(x, x, a)
permute₃(x, y, a) -> permute₃(isZero₁(x), x, b)
permute₃(false, x, b) -> permute₃(ack₂(x, x), p₁(x), c)
permute₃(true, x, b) -> 0
permute₃(y, x, c) -> s₁(s₁(permute₃(x, y, a)))
p₁(0) -> 0
p₁(s₁(x)) -> x
ack₂(0, x) -> plus₂(x, s₁(0))
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(x, s₁(0)) -> s₁(x)
plus₂(x, 0) -> x
isZero₁(0) -> true
isZero₁(s₁(x)) -> false

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

double₁(x) -> permute₃(x, x, a)
permute₃(x, y, a) -> permute₃(isZero₁(x), x, b)
permute₃(false, x, b) -> permute₃(ack₂(x, x), p₁(x), c)
permute₃(true, x, b) -> 0
permute₃(y, x, c) -> s₁(s₁(permute₃(x, y, a)))
p₁(0) -> 0
p₁(s₁(x)) -> x
ack₂(0, x) -> plus₂(x, s₁(0))
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(x, s₁(0)) -> s₁(x)
plus₂(x, 0) -> x
isZero₁(0) -> true
isZero₁(s₁(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:

PERMUTE₃(x, y, a) -> ISZERO₁(x)
PLUS₂(x, s₁(s₁(y))) -> PLUS₂(s₁(x), y)
PLUS₂(s₁(x), y) -> PLUS₂(x, s₁(y))
PERMUTE₃(false, x, b) -> PERMUTE₃(ack₂(x, x), p₁(x), c)
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))
ACK₂(0, x) -> PLUS₂(x, s₁(0))
PERMUTE₃(x, y, a) -> PERMUTE₃(isZero₁(x), x, b)
ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)
DOUBLE₁(x) -> PERMUTE₃(x, x, a)
ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
PERMUTE₃(false, x, b) -> ACK₂(x, x)
PERMUTE₃(y, x, c) -> PERMUTE₃(x, y, a)
PERMUTE₃(false, x, b) -> P₁(x)

The TRS R consists of the following rules:

double₁(x) -> permute₃(x, x, a)
permute₃(x, y, a) -> permute₃(isZero₁(x), x, b)
permute₃(false, x, b) -> permute₃(ack₂(x, x), p₁(x), c)
permute₃(true, x, b) -> 0
permute₃(y, x, c) -> s₁(s₁(permute₃(x, y, a)))
p₁(0) -> 0
p₁(s₁(x)) -> x
ack₂(0, x) -> plus₂(x, s₁(0))
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(x, s₁(0)) -> s₁(x)
plus₂(x, 0) -> x
isZero₁(0) -> true
isZero₁(s₁(x)) -> false

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:

PERMUTE₃(x, y, a) -> ISZERO₁(x)
PLUS₂(x, s₁(s₁(y))) -> PLUS₂(s₁(x), y)
PLUS₂(s₁(x), y) -> PLUS₂(x, s₁(y))
PERMUTE₃(false, x, b) -> PERMUTE₃(ack₂(x, x), p₁(x), c)
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))
ACK₂(0, x) -> PLUS₂(x, s₁(0))
PERMUTE₃(x, y, a) -> PERMUTE₃(isZero₁(x), x, b)
ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)
DOUBLE₁(x) -> PERMUTE₃(x, x, a)
ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
PERMUTE₃(false, x, b) -> ACK₂(x, x)
PERMUTE₃(y, x, c) -> PERMUTE₃(x, y, a)
PERMUTE₃(false, x, b) -> P₁(x)

The TRS R consists of the following rules:

double₁(x) -> permute₃(x, x, a)
permute₃(x, y, a) -> permute₃(isZero₁(x), x, b)
permute₃(false, x, b) -> permute₃(ack₂(x, x), p₁(x), c)
permute₃(true, x, b) -> 0
permute₃(y, x, c) -> s₁(s₁(permute₃(x, y, a)))
p₁(0) -> 0
p₁(s₁(x)) -> x
ack₂(0, x) -> plus₂(x, s₁(0))
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(x, s₁(0)) -> s₁(x)
plus₂(x, 0) -> x
isZero₁(0) -> true
isZero₁(s₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PLUS₂(x, s₁(s₁(y))) -> PLUS₂(s₁(x), y)
PLUS₂(s₁(x), y) -> PLUS₂(x, s₁(y))

The TRS R consists of the following rules:

double₁(x) -> permute₃(x, x, a)
permute₃(x, y, a) -> permute₃(isZero₁(x), x, b)
permute₃(false, x, b) -> permute₃(ack₂(x, x), p₁(x), c)
permute₃(true, x, b) -> 0
permute₃(y, x, c) -> s₁(s₁(permute₃(x, y, a)))
p₁(0) -> 0
p₁(s₁(x)) -> x
ack₂(0, x) -> plus₂(x, s₁(0))
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(x, s₁(0)) -> s₁(x)
plus₂(x, 0) -> x
isZero₁(0) -> true
isZero₁(s₁(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.

PLUS₂(x, s₁(s₁(y))) -> PLUS₂(s₁(x), y)

The remaining pairs can at least be oriented weakly.

PLUS₂(s₁(x), y) -> PLUS₂(x, s₁(y))

Used ordering: Polynomial interpretation [21]:

POL(PLUS₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PLUS₂(s₁(x), y) -> PLUS₂(x, s₁(y))

The TRS R consists of the following rules:

double₁(x) -> permute₃(x, x, a)
permute₃(x, y, a) -> permute₃(isZero₁(x), x, b)
permute₃(false, x, b) -> permute₃(ack₂(x, x), p₁(x), c)
permute₃(true, x, b) -> 0
permute₃(y, x, c) -> s₁(s₁(permute₃(x, y, a)))
p₁(0) -> 0
p₁(s₁(x)) -> x
ack₂(0, x) -> plus₂(x, s₁(0))
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(x, s₁(0)) -> s₁(x)
plus₂(x, 0) -> x
isZero₁(0) -> true
isZero₁(s₁(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.

PLUS₂(s₁(x), y) -> PLUS₂(x, s₁(y))

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

POL(PLUS₂(x₁, x₂)) = 2·x₁ + x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

double₁(x) -> permute₃(x, x, a)
permute₃(x, y, a) -> permute₃(isZero₁(x), x, b)
permute₃(false, x, b) -> permute₃(ack₂(x, x), p₁(x), c)
permute₃(true, x, b) -> 0
permute₃(y, x, c) -> s₁(s₁(permute₃(x, y, a)))
p₁(0) -> 0
p₁(s₁(x)) -> x
ack₂(0, x) -> plus₂(x, s₁(0))
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(x, s₁(0)) -> s₁(x)
plus₂(x, 0) -> x
isZero₁(0) -> true
isZero₁(s₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)
ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The TRS R consists of the following rules:

double₁(x) -> permute₃(x, x, a)
permute₃(x, y, a) -> permute₃(isZero₁(x), x, b)
permute₃(false, x, b) -> permute₃(ack₂(x, x), p₁(x), c)
permute₃(true, x, b) -> 0
permute₃(y, x, c) -> s₁(s₁(permute₃(x, y, a)))
p₁(0) -> 0
p₁(s₁(x)) -> x
ack₂(0, x) -> plus₂(x, s₁(0))
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(x, s₁(0)) -> s₁(x)
plus₂(x, 0) -> x
isZero₁(0) -> true
isZero₁(s₁(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.

ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The remaining pairs can at least be oriented weakly.

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)

Used ordering: Polynomial interpretation [21]:

POL(0) = 2
POL(ACK₂(x₁, x₂)) = x₁
POL(ack₂(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(plus₂(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)

The TRS R consists of the following rules:

double₁(x) -> permute₃(x, x, a)
permute₃(x, y, a) -> permute₃(isZero₁(x), x, b)
permute₃(false, x, b) -> permute₃(ack₂(x, x), p₁(x), c)
permute₃(true, x, b) -> 0
permute₃(y, x, c) -> s₁(s₁(permute₃(x, y, a)))
p₁(0) -> 0
p₁(s₁(x)) -> x
ack₂(0, x) -> plus₂(x, s₁(0))
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(x, s₁(0)) -> s₁(x)
plus₂(x, 0) -> x
isZero₁(0) -> true
isZero₁(s₁(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.

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)

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

POL(ACK₂(x₁, x₂)) = 2·x₂
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

double₁(x) -> permute₃(x, x, a)
permute₃(x, y, a) -> permute₃(isZero₁(x), x, b)
permute₃(false, x, b) -> permute₃(ack₂(x, x), p₁(x), c)
permute₃(true, x, b) -> 0
permute₃(y, x, c) -> s₁(s₁(permute₃(x, y, a)))
p₁(0) -> 0
p₁(s₁(x)) -> x
ack₂(0, x) -> plus₂(x, s₁(0))
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(x, s₁(0)) -> s₁(x)
plus₂(x, 0) -> x
isZero₁(0) -> true
isZero₁(s₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PERMUTE₃(y, x, c) -> PERMUTE₃(x, y, a)
PERMUTE₃(false, x, b) -> PERMUTE₃(ack₂(x, x), p₁(x), c)
PERMUTE₃(x, y, a) -> PERMUTE₃(isZero₁(x), x, b)

The TRS R consists of the following rules:

double₁(x) -> permute₃(x, x, a)
permute₃(x, y, a) -> permute₃(isZero₁(x), x, b)
permute₃(false, x, b) -> permute₃(ack₂(x, x), p₁(x), c)
permute₃(true, x, b) -> 0
permute₃(y, x, c) -> s₁(s₁(permute₃(x, y, a)))
p₁(0) -> 0
p₁(s₁(x)) -> x
ack₂(0, x) -> plus₂(x, s₁(0))
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> plus₂(x, s₁(y))
plus₂(x, s₁(s₁(y))) -> s₁(plus₂(s₁(x), y))
plus₂(x, s₁(0)) -> s₁(x)
plus₂(x, 0) -> x
isZero₁(0) -> true
isZero₁(s₁(x)) -> false

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