Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(6) Obligation:

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.

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:

s₁ > ACKIN₂

Status:

ACKIN₂: [1,2]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE