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:

ackin(s(X), s(Y)) → u21(ackin(s(X), Y), X)
u21(ackout(X), Y) → u22(ackin(Y, X))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

ackin(s(X), s(Y)) → u21(ackin(s(X), Y), X)
u21(ackout(X), Y) → u22(ackin(Y, X))

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:

ackin(s(X), s(Y)) → u21(ackin(s(X), Y), X)
u21(ackout(X), Y) → u22(ackin(Y, X))

The set Q consists of the following terms:

ackin(s(x0), s(x1))
u21(ackout(x0), x1)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

U21(ackout(X), Y) → ACKIN(Y, X)
ACKIN(s(X), s(Y)) → ACKIN(s(X), Y)
ACKIN(s(X), s(Y)) → U21(ackin(s(X), Y), X)

The TRS R consists of the following rules:

ackin(s(X), s(Y)) → u21(ackin(s(X), Y), X)
u21(ackout(X), Y) → u22(ackin(Y, X))

The set Q consists of the following terms:

ackin(s(x0), s(x1))
u21(ackout(x0), x1)

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:

U21(ackout(X), Y) → ACKIN(Y, X)
ACKIN(s(X), s(Y)) → ACKIN(s(X), Y)
ACKIN(s(X), s(Y)) → U21(ackin(s(X), Y), X)

The TRS R consists of the following rules:

ackin(s(X), s(Y)) → u21(ackin(s(X), Y), X)
u21(ackout(X), Y) → u22(ackin(Y, X))

The set Q consists of the following terms:

ackin(s(x0), s(x1))
u21(ackout(x0), x1)

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:

ACKIN(s(X), s(Y)) → ACKIN(s(X), Y)
U21(ackout(X), Y) → ACKIN(Y, X)
ACKIN(s(X), s(Y)) → U21(ackin(s(X), Y), X)

The TRS R consists of the following rules:

ackin(s(X), s(Y)) → u21(ackin(s(X), Y), X)
u21(ackout(X), Y) → u22(ackin(Y, X))

The set Q consists of the following terms:

ackin(s(x0), s(x1))
u21(ackout(x0), x1)

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:

ACKIN(s(X), s(Y)) → ACKIN(s(X), Y)

The TRS R consists of the following rules:

ackin(s(X), s(Y)) → u21(ackin(s(X), Y), X)
u21(ackout(X), Y) → u22(ackin(Y, X))

The set Q consists of the following terms:

ackin(s(x0), s(x1))
u21(ackout(x0), x1)

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.


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

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
QDP
                      ↳ PisEmptyProof

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

ackin(s(X), s(Y)) → u21(ackin(s(X), Y), X)
u21(ackout(X), Y) → u22(ackin(Y, X))

The set Q consists of the following terms:

ackin(s(x0), s(x1))
u21(ackout(x0), x1)

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