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. lexicographic path order with precedence:
    
      precedence:
      
        ack_in > s > u21 > u11 > u22 > ack_in# > ack_out > u21# > |0|
      
      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) = [1]
        pi(u21) = [1]
    
    2. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        ack_in#_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,1,0,0)) x1 + (0,1,1,0)
        s_A(x1) = x1 + (2,1,1,1)
        |0|_A() = (1,1,1,10)
        u21#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,1,1,0),(1,0,0,0)) x2 + (3,0,0,2)
        ack_in_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,0,0),(0,0,0,0)) x2 + (1,1,1,1)
        ack_out_A(x1) = (3,3,5,11)
        u22_A(x1) = (7,3,5,11)
        u11_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,1,1,0)) x1 + (1,0,5,0)
        u21_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x1 + (3,4,4,0)
    

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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > ack_in#
      
      argument filter:
    
        pi(ack_in#) = [2]
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        ack_in#_A(x1,x2) = (0,0,0,0)
        s_A(x1) = (1,1,1,1)
    

The next rules are strictly ordered:

  p1

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