YES

We show the termination of the TRS R:

  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)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: ack_in#(s(m),|0|()) -> u11#(ack_in(m,s(|0|())))
p2: ack_in#(s(m),|0|()) -> ack_in#(m,s(|0|()))
p3: ack_in#(s(m),s(n)) -> u21#(ack_in(s(m),n),m)
p4: ack_in#(s(m),s(n)) -> ack_in#(s(m),n)
p5: u21#(ack_out(n),m) -> u22#(ack_in(m,n))
p6: u21#(ack_out(n),m) -> ack_in#(m,n)

and R consists of:

r1: ack_in(|0|(),n) -> ack_out(s(n))
r2: ack_in(s(m),|0|()) -> u11(ack_in(m,s(|0|())))
r3: u11(ack_out(n)) -> ack_out(n)
r4: ack_in(s(m),s(n)) -> u21(ack_in(s(m),n),m)
r5: u21(ack_out(n),m) -> u22(ack_in(m,n))
r6: u22(ack_out(n)) -> ack_out(n)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p6}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: ack_in#(s(m),|0|()) -> ack_in#(m,s(|0|()))
p2: ack_in#(s(m),s(n)) -> ack_in#(s(m),n)
p3: ack_in#(s(m),s(n)) -> u21#(ack_in(s(m),n),m)
p4: u21#(ack_out(n),m) -> ack_in#(m,n)

and R consists of:

r1: ack_in(|0|(),n) -> ack_out(s(n))
r2: ack_in(s(m),|0|()) -> u11(ack_in(m,s(|0|())))
r3: u11(ack_out(n)) -> ack_out(n)
r4: ack_in(s(m),s(n)) -> u21(ack_in(s(m),n),m)
r5: u21(ack_out(n),m) -> u22(ack_in(m,n))
r6: u22(ack_out(n)) -> ack_out(n)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      ack_in > u21 > u22 > |0| > u11 > ack_out > ack_in# > s > u21#
    
    argument filter:
  
      pi(ack_in#) = 1
      pi(s) = [1]
      pi(|0|) = []
      pi(u21#) = [2]
      pi(ack_in) = [1, 2]
      pi(ack_out) = []
      pi(u22) = []
      pi(u11) = []
      pi(u21) = []

The next rules are strictly ordered:

  p1, p3, p4

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: ack_in#(s(m),s(n)) -> ack_in#(s(m),n)

and R consists of:

r1: ack_in(|0|(),n) -> ack_out(s(n))
r2: ack_in(s(m),|0|()) -> u11(ack_in(m,s(|0|())))
r3: u11(ack_out(n)) -> ack_out(n)
r4: ack_in(s(m),s(n)) -> u21(ack_in(s(m),n),m)
r5: u21(ack_out(n),m) -> u22(ack_in(m,n))
r6: u22(ack_out(n)) -> ack_out(n)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: ack_in#(s(m),s(n)) -> ack_in#(s(m),n)

and R consists of:

r1: ack_in(|0|(),n) -> ack_out(s(n))
r2: ack_in(s(m),|0|()) -> u11(ack_in(m,s(|0|())))
r3: u11(ack_out(n)) -> ack_out(n)
r4: ack_in(s(m),s(n)) -> u21(ack_in(s(m),n),m)
r5: u21(ack_out(n),m) -> u22(ack_in(m,n))
r6: u22(ack_out(n)) -> ack_out(n)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      s > ack_in#
    
    argument filter:
  
      pi(ack_in#) = [2]
      pi(s) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.