Termination w.r.t. Q proof of TRS/AProVErta1.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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:

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

The TRS R consists of the following rules:

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

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, plus(y, 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) = ACK(x1)
s(x1) = s(x1)
plus(x1, x2) = plus(x1, x2)
ack(x1, x2) = ack
0 = 0

Lexicographic path order with status [19].
Precedence:

ack > s₁ > plus₂
0 > ACK₁ > plus₂
0 > s₁ > plus₂

Status:

plus₂: [2,1]
s₁: [1]
0: multiset
ACK₁: [1]
ack: []

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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:

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

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)

Lexicographic path order with status [19].
Precedence:

trivial

Status:

s₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

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

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.