Termination w.r.t. Q proof of TRS/Cimeack

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:

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

The set Q consists of the following terms:

ack_in(0, x0)
ack_in(s(x0), 0)
u11(ack_out(x0))
ack_in(s(x0), s(x1))
u21(ack_out(x0), x1)
u22(ack_out(x0))

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_IN(s(m), s(n)) → ACK_IN(s(m), n)
ACK_IN(s(m), 0) → U11(ack_in(m, s(0)))
ACK_IN(s(m), 0) → ACK_IN(m, s(0))
U21(ack_out(n), m) → ACK_IN(m, n)
ACK_IN(s(m), s(n)) → U21(ack_in(s(m), n), m)
U21(ack_out(n), m) → U22(ack_in(m, n))

The TRS R consists of the following rules:

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

The set Q consists of the following terms:

ack_in(0, x0)
ack_in(s(x0), 0)
u11(ack_out(x0))
ack_in(s(x0), s(x1))
u21(ack_out(x0), x1)
u22(ack_out(x0))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

ACK_IN(s(m), s(n)) → ACK_IN(s(m), n)
ACK_IN(s(m), 0) → U11(ack_in(m, s(0)))
ACK_IN(s(m), 0) → ACK_IN(m, s(0))
U21(ack_out(n), m) → ACK_IN(m, n)
ACK_IN(s(m), s(n)) → U21(ack_in(s(m), n), m)
U21(ack_out(n), m) → U22(ack_in(m, n))

The TRS R consists of the following rules:

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

The set Q consists of the following terms:

ack_in(0, x0)
ack_in(s(x0), 0)
u11(ack_out(x0))
ack_in(s(x0), s(x1))
u21(ack_out(x0), x1)
u22(ack_out(x0))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

ACK_IN(s(m), s(n)) → ACK_IN(s(m), n)
ACK_IN(s(m), 0) → U11(ack_in(m, s(0)))
ACK_IN(s(m), 0) → ACK_IN(m, s(0))
ACK_IN(s(m), s(n)) → U21(ack_in(s(m), n), m)
U21(ack_out(n), m) → ACK_IN(m, n)
U21(ack_out(n), m) → U22(ack_in(m, n))

The TRS R consists of the following rules:

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

The set Q consists of the following terms:

ack_in(0, x0)
ack_in(s(x0), 0)
u11(ack_out(x0))
ack_in(s(x0), s(x1))
u21(ack_out(x0), x1)
u22(ack_out(x0))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

ACK_IN(s(m), s(n)) → ACK_IN(s(m), n)
ACK_IN(s(m), 0) → ACK_IN(m, s(0))
U21(ack_out(n), m) → ACK_IN(m, n)
ACK_IN(s(m), s(n)) → U21(ack_in(s(m), n), m)

The TRS R consists of the following rules:

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

The set Q consists of the following terms:

ack_in(0, x0)
ack_in(s(x0), 0)
u11(ack_out(x0))
ack_in(s(x0), s(x1))
u21(ack_out(x0), x1)
u22(ack_out(x0))

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_IN(s(m), 0) → ACK_IN(m, s(0))
ACK_IN(s(m), s(n)) → U21(ack_in(s(m), n), m)

The remaining pairs can at least be oriented weakly.

ACK_IN(s(m), s(n)) → ACK_IN(s(m), n)
U21(ack_out(n), m) → ACK_IN(m, n)

Used ordering: Combined order from the following AFS and order.
ACK_IN(x1, x2) = x1
s(x1) = s(x1)
U21(x1, x2) = x2

Recursive path order with status [2].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

ACK_IN(s(m), s(n)) → ACK_IN(s(m), n)
U21(ack_out(n), m) → ACK_IN(m, n)

The TRS R consists of the following rules:

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

The set Q consists of the following terms:

ack_in(0, x0)
ack_in(s(x0), 0)
u11(ack_out(x0))
ack_in(s(x0), s(x1))
u21(ack_out(x0), x1)
u22(ack_out(x0))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

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

ACK_IN(s(m), s(n)) → ACK_IN(s(m), n)

The TRS R consists of the following rules:

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

The set Q consists of the following terms:

ack_in(0, x0)
ack_in(s(x0), 0)
u11(ack_out(x0))
ack_in(s(x0), s(x1))
u21(ack_out(x0), x1)
u22(ack_out(x0))

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_IN(s(m), s(n)) → ACK_IN(s(m), n)

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

Recursive path order with status [2].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ PisEmptyProof

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

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

The set Q consists of the following terms:

ack_in(0, x0)
ack_in(s(x0), 0)
u11(ack_out(x0))
ack_in(s(x0), s(x1))
u21(ack_out(x0), x1)
u22(ack_out(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.