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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ack_in#_A(x1,x2) = max{0, x1 - 28, x2 - 17}
            s_A(x1) = max{4, x1}
            |0|_A = 3
            u21#_A(x1,x2) = max{0, x1 - 12, x2 - 28}
            ack_in_A(x1,x2) = max{11, x1 - 19, x2 - 5}
            ack_out_A(x1) = max{1, x1 - 5}
            u11_A(x1) = x1
            u21_A(x1,x2) = max{11, x1, x2 - 19}
            u22_A(x1) = max{2, x1}
            ack_in#!_A(x1,x2) = max{9, x1 - 20, x2 - 7}
            s!_A(x1) = max{10, x1 - 18}
            |0|!_A = 2
            u21#!_A(x1,x2) = max{9, x1 - 2, x2 - 20}
            ack_in!_A(x1,x2) = max{10, x1 - 13, x2 - 8}
            ack_out!_A(x1) = max{3, x1 - 32}
            u11!_A(x1) = max{1, x1 - 28}
            u21!_A(x1,x2) = max{5, x1 - 14, x2 - 14}
            u22!_A(x1) = max{4, x1 - 26}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ack_in#_A(x1,x2) = max{0, x1 - 40, x2 - 42}
            s_A(x1) = max{24, x1}
            |0|_A = 27
            u21#_A(x1,x2) = max{0, x1 - 12, x2 - 40}
            ack_in_A(x1,x2) = max{11, x1 - 28, x2 - 30}
            ack_out_A(x1) = max{1, x1 - 30}
            u11_A(x1) = x1
            u21_A(x1,x2) = max{11, x1, x2 - 28}
            u22_A(x1) = x1
            ack_in#!_A(x1,x2) = max{6, x1 - 21, x2 - 44}
            s!_A(x1) = max{2, x1 - 56}
            |0|!_A = 4
            u21#!_A(x1,x2) = max{6, x1 - 14, x2 - 21}
            ack_in!_A(x1,x2) = max{8, x1 - 28, x2 - 32}
            ack_out!_A(x1) = max{5, x1 - 43}
            u11!_A(x1) = max{1, x1 - 51}
            u21!_A(x1,x2) = max{7, x1 - 3, x2 - 28}
            u22!_A(x1) = max{6, x1 - 12}
        
  
    partial status:
  
      pi(ack_in#) = [1]
      pi(s) = [1]
      pi(|0|) = []
      pi(u21#) = [2]
      pi(ack_in) = [1, 2]
      pi(ack_out) = []
      pi(u11) = [1]
      pi(u21) = []
      pi(u22) = []

The following rules are removed:

  p1, p3



-- 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)
p2: 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:

  {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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ack_in#_A(x1,x2) = 0
            s_A(x1) = max{5, x1 - 19}
            ack_in#!_A(x1,x2) = 0
            s!_A(x1) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ack_in#_A(x1,x2) = x2
            s_A(x1) = max{26, x1 + 18}
            ack_in#!_A(x1,x2) = max{4, x2 + 1}
            s!_A(x1) = max{3, x1 - 72}
        
  
    partial status:
  
      pi(ack_in#) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.