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:

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.

(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:

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))

(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:

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

(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:

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACK_IN(x1, x2)  =  x1
s(x1)  =  s(x1)
U21(x1, x2)  =  U21(x2)
ack_in(x1, x2)  =  ack_in
ack_out(x1)  =  ack_out
0  =  0
u11(x1)  =  u11
u21(x1, x2)  =  u21
u22(x1)  =  u22

Lexicographic path order with status [LPO].
Precedence:

s₁ > U21₁ > u11
s₁ > u21 > ack_in > u11
0 > ack_in > u11
0 > ack_out > u11
u22 > ack_out > u11

Status:

ack_in: []
u22: []
u11: []
u21: []
s₁: [1]
0: []
ack_out: []
U21₁: [1]

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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.

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

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.

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) TRUE