YES
We show the termination of the TRS R:

  plus(s(s(x)),y) -> s(plus(x,s(y)))
  plus(x,s(s(y))) -> s(plus(s(x),y))
  plus(s(|0|()),y) -> s(y)
  plus(|0|(),y) -> y
  ack(|0|(),y) -> s(y)
  ack(s(x),|0|()) -> ack(x,s(|0|()))
  ack(s(x),s(y)) -> ack(x,plus(y,ack(s(x),y)))


-- SCC decomposition.

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

p1: plus#(s(s(x)),y) -> plus#(x,s(y))
p2: plus#(x,s(s(y))) -> plus#(s(x),y)
p3: ack#(s(x),|0|()) -> ack#(x,s(|0|()))
p4: ack#(s(x),s(y)) -> ack#(x,plus(y,ack(s(x),y)))
p5: ack#(s(x),s(y)) -> plus#(y,ack(s(x),y))
p6: ack#(s(x),s(y)) -> ack#(s(x),y)

and R consists of:

r1: plus(s(s(x)),y) -> s(plus(x,s(y)))
r2: plus(x,s(s(y))) -> s(plus(s(x),y))
r3: plus(s(|0|()),y) -> s(y)
r4: plus(|0|(),y) -> y
r5: ack(|0|(),y) -> s(y)
r6: ack(s(x),|0|()) -> ack(x,s(|0|()))
r7: ack(s(x),s(y)) -> ack(x,plus(y,ack(s(x),y)))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p6}
  {p1, p2}


-- Reduction pair.

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

p1: ack#(s(x),s(y)) -> ack#(s(x),y)
p2: ack#(s(x),s(y)) -> ack#(x,plus(y,ack(s(x),y)))
p3: ack#(s(x),|0|()) -> ack#(x,s(|0|()))

and R consists of:

r1: plus(s(s(x)),y) -> s(plus(x,s(y)))
r2: plus(x,s(s(y))) -> s(plus(s(x),y))
r3: plus(s(|0|()),y) -> s(y)
r4: plus(|0|(),y) -> y
r5: ack(|0|(),y) -> s(y)
r6: ack(s(x),|0|()) -> ack(x,s(|0|()))
r7: ack(s(x),s(y)) -> ack(x,plus(y,ack(s(x),y)))

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ack#_A(x1,x2) = 0
              s_A(x1) = 2
              plus_A(x1,x2) = x1 + 1
              ack_A(x1,x2) = 1
              |0|_A() = 1
              ack#!_A(x1,x2) = 0
              s!_A(x1) = 3
              plus!_A(x1,x2) = 4
              ack!_A(x1,x2) = 2
              |0|!_A() = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ack#_A(x1,x2) = x1
              s_A(x1) = x1 + 2
              plus_A(x1,x2) = x2
              ack_A(x1,x2) = x1 + x2 + 3
              |0|_A() = 1
              ack#!_A(x1,x2) = 1
              s!_A(x1) = 3
              plus!_A(x1,x2) = x1 + 2
              ack!_A(x1,x2) = x2 + 4
              |0|!_A() = 0
        
  
    partial status:
  
      pi(ack#) = [1]
      pi(s) = [1]
      pi(plus) = [1]
      pi(ack) = [1, 2]
      pi(|0|) = []

The following rules are removed:

  p2, p3



-- SCC decomposition.

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

p1: ack#(s(x),s(y)) -> ack#(s(x),y)

and R consists of:

r1: plus(s(s(x)),y) -> s(plus(x,s(y)))
r2: plus(x,s(s(y))) -> s(plus(s(x),y))
r3: plus(s(|0|()),y) -> s(y)
r4: plus(|0|(),y) -> y
r5: ack(|0|(),y) -> s(y)
r6: ack(s(x),|0|()) -> ack(x,s(|0|()))
r7: ack(s(x),s(y)) -> ack(x,plus(y,ack(s(x),y)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: ack#(s(x),s(y)) -> ack#(s(x),y)

and R consists of:

r1: plus(s(s(x)),y) -> s(plus(x,s(y)))
r2: plus(x,s(s(y))) -> s(plus(s(x),y))
r3: plus(s(|0|()),y) -> s(y)
r4: plus(|0|(),y) -> y
r5: ack(|0|(),y) -> s(y)
r6: ack(s(x),|0|()) -> ack(x,s(|0|()))
r7: ack(s(x),s(y)) -> ack(x,plus(y,ack(s(x),y)))

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ack#_A(x1,x2) = x2
              s_A(x1) = x1 + 2
              ack#!_A(x1,x2) = 1
              s!_A(x1) = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ack#_A(x1,x2) = x2
              s_A(x1) = x1 + 4
              ack#!_A(x1,x2) = x1 + x2
              s!_A(x1) = 5
        
  
    partial status:
  
      pi(ack#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: plus#(s(s(x)),y) -> plus#(x,s(y))
p2: plus#(x,s(s(y))) -> plus#(s(x),y)

and R consists of:

r1: plus(s(s(x)),y) -> s(plus(x,s(y)))
r2: plus(x,s(s(y))) -> s(plus(s(x),y))
r3: plus(s(|0|()),y) -> s(y)
r4: plus(|0|(),y) -> y
r5: ack(|0|(),y) -> s(y)
r6: ack(s(x),|0|()) -> ack(x,s(|0|()))
r7: ack(s(x),s(y)) -> ack(x,plus(y,ack(s(x),y)))

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = 0
              s_A(x1) = x1 + 2
              plus#!_A(x1,x2) = x1 + x2 + 2
              s!_A(x1) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = 0
              s_A(x1) = x1 + 4
              plus#!_A(x1,x2) = 2
              s!_A(x1) = 2
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.