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 combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        ack_in#_A(x1,x2) = ((1,1),(0,0)) x1
        s_A(x1) = ((1,1),(0,0)) x1 + (0,2)
        |0|_A() = (1,0)
        u21#_A(x1,x2) = x1 + ((1,1),(0,0)) x2
        ack_in_A(x1,x2) = x1 + ((0,0),(1,1)) x2 + (2,1)
        ack_out_A(x1) = (0,1)
        u22_A(x1) = (1,1)
        u11_A(x1) = (1,1)
        u21_A(x1,x2) = (2,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        ack_in#_A(x1,x2) = ((1,1),(0,1)) x1 + (0,2)
        s_A(x1) = ((0,0),(1,1)) x1 + (1,2)
        |0|_A() = (1,1)
        u21#_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,1),(0,1)) x2 + (1,0)
        ack_in_A(x1,x2) = (4,0)
        ack_out_A(x1) = (2,2)
        u22_A(x1) = (1,1)
        u11_A(x1) = (3,1)
        u21_A(x1,x2) = (0,0)
    

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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        ack_in#_A(x1,x2) = x1 + ((1,1),(1,0)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        ack_in#_A(x1,x2) = x1 + ((1,1),(1,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6

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