Termination w.r.t. Q proof of /tmp/tmpxr49BU/rta1.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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

      ↳ DependencyGraphProof

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

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

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 [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

PLUS(s(s(x)), y) → PLUS(x, s(y))
PLUS(x, s(s(y))) → PLUS(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 [15].

The following pairs can be oriented strictly and are deleted.

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

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

POL(PLUS(x₁, x₂)) = (4)x_1 + (4)x_2
POL(s(x₁)) = 4 + x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), s(y)) → ACK(x, plus(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 [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

ACK(s(x), s(y)) → ACK(s(x), y)

Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = 2
POL(ACK(x₁, x₂)) = (3)x_1
POL(s(x₁)) = 4 + (2)x_1
POL(0) = 0
POL(ack(x₁, x₂)) = 0

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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 [15].

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: Polynomial interpretation [25,35]:

POL(ACK(x₁, x₂)) = (4)x_2
POL(s(x₁)) = 1 + (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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.