YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: ack(|0|(),y) -> s(y)
r2: ack(s(x),|0|()) -> ack(x,s(|0|()))
r3: ack(s(x),s(y)) -> ack(x,ack(s(x),y))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

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

and R consists of:

r1: ack(|0|(),y) -> s(y)
r2: ack(s(x),|0|()) -> ack(x,s(|0|()))
r3: ack(s(x),s(y)) -> ack(x,ack(s(x),y))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > ack# > ack > s
      
      argument filter:
    
        pi(ack#) = [1, 2]
        pi(s) = [1]
        pi(|0|) = []
        pi(ack) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > ack > ack# > s
      
      argument filter:
    
        pi(ack#) = []
        pi(s) = []
        pi(|0|) = []
        pi(ack) = 2
    

The next rules are strictly ordered:

  p1, p2, p3

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