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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        ack#_A(x1,x2) = max{x1 + 11, x2 + 7}
        s_A(x1) = max{2, x1}
        |0|_A = 3
        ack_A(x1,x2) = max{x1 + 4, x2}
  
    precedence:
    
      ack > ack# = s = |0|
    
    partial status:
  
      pi(ack#) = [1]
      pi(s) = [1]
      pi(|0|) = []
      pi(ack) = [1, 2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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,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}


-- 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,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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        ack#_A(x1,x2) = max{4, x1 + 1}
        s_A(x1) = max{2, x1}
        ack_A(x1,x2) = max{x1 + 3, x2}
        |0|_A = 1
  
    precedence:
    
      ack = |0| > ack# = s
    
    partial status:
  
      pi(ack#) = [1]
      pi(s) = [1]
      pi(ack) = [1, 2]
      pi(|0|) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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: 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}


-- 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: 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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        ack#_A(x1,x2) = max{0, x2 - 2}
        s_A(x1) = max{4, x1 + 1}
  
    precedence:
    
      ack# = s
    
    partial status:
  
      pi(ack#) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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