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) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:

[ack_in2, u21₂] > ack_out1
[ack_in2, u21₂] > s₁
[ack_in2, u21₂] > u11₁
[ack_in2, u21₂] > u22₁

Status:

ack_in2: [1,2]
u21₂: [2,1]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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)

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE