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))
  f(s(x),y) -> f(x,s(x))
  f(x,s(y)) -> f(y,x)
  f(x,y) -> ack(x,y)
  ack(s(x),y) -> f(x,x)


-- 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)
p4: f#(s(x),y) -> f#(x,s(x))
p5: f#(x,s(y)) -> f#(y,x)
p6: f#(x,y) -> ack#(x,y)
p7: ack#(s(x),y) -> f#(x,x)

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))
r4: f(s(x),y) -> f(x,s(x))
r5: f(x,s(y)) -> f(y,x)
r6: f(x,y) -> ack(x,y)
r7: ack(s(x),y) -> f(x,x)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- 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),y) -> f#(x,x)
p3: f#(x,y) -> ack#(x,y)
p4: ack#(s(x),s(y)) -> ack#(s(x),y)
p5: ack#(s(x),s(y)) -> ack#(x,ack(s(x),y))
p6: f#(x,s(y)) -> f#(y,x)
p7: f#(s(x),y) -> f#(x,s(x))

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))
r4: f(s(x),y) -> f(x,s(x))
r5: f(x,s(y)) -> f(y,x)
r6: f(x,y) -> ack(x,y)
r7: ack(s(x),y) -> f(x,x)

The set of usable rules consists of

  r1

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#_A(x1,x2) = max{1, x1 - 2}
            s_A(x1) = max{1, x1}
            |0|_A = 4
            f#_A(x1,x2) = max{1, x1 - 2, x2 - 2}
            ack_A(x1,x2) = max{2, x1, x2}
            ack#!_A(x1,x2) = 1
            s!_A(x1) = max{10, x1 + 2}
            |0|!_A = 12
            f#!_A(x1,x2) = 1
            ack!_A(x1,x2) = max{13, x1 + 7, x2 + 7}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ack#_A(x1,x2) = 1
            s_A(x1) = max{6, x1 + 4}
            |0|_A = 8
            f#_A(x1,x2) = 1
            ack_A(x1,x2) = max{15, x2 + 5}
            ack#!_A(x1,x2) = max{6, x1 + 2}
            s!_A(x1) = x1 + 11
            |0|!_A = 19
            f#!_A(x1,x2) = max{x1 + 6, x2 + 3}
            ack!_A(x1,x2) = max{x1 + 26, x2 + 17}
        
  
    partial status:
  
      pi(ack#) = []
      pi(s) = []
      pi(|0|) = []
      pi(f#) = []
      pi(ack) = [2]

The following rules are removed:

  p1, p5, p6, p7



-- SCC decomposition.

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

p1: ack#(s(x),y) -> f#(x,x)
p2: f#(x,y) -> ack#(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))
r4: f(s(x),y) -> f(x,s(x))
r5: f(x,s(y)) -> f(y,x)
r6: f(x,y) -> ack(x,y)
r7: ack(s(x),y) -> f(x,x)

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),y) -> f#(x,x)
p2: f#(x,y) -> ack#(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))
r4: f(s(x),y) -> f(x,s(x))
r5: f(x,s(y)) -> f(y,x)
r6: f(x,y) -> ack(x,y)
r7: ack(s(x),y) -> f(x,x)

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#_A(x1,x2) = max{0, x1 - 3, x2 - 3}
            s_A(x1) = max{2, x1}
            f#_A(x1,x2) = max{0, x1 - 3, x2 - 3}
            ack#!_A(x1,x2) = max{1, x1 - 14, x2 - 16}
            s!_A(x1) = max{2, x1 - 11}
            f#!_A(x1,x2) = max{1, x1 - 14, x2 - 15}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ack#_A(x1,x2) = max{0, x1 - 5, x2 - 4}
            s_A(x1) = max{4, x1 + 1}
            f#_A(x1,x2) = max{0, x1 - 4, x2 - 4}
            ack#!_A(x1,x2) = max{1, x1 - 22, x2 - 21}
            s!_A(x1) = max{2, x1 - 17}
            f#!_A(x1,x2) = max{1, x1 - 22, x2 - 21}
        
  
    partial status:
  
      pi(ack#) = [1]
      pi(s) = [1]
      pi(f#) = [1]

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: f#(x,y) -> ack#(x,y)
p2: 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))
r4: f(s(x),y) -> f(x,s(x))
r5: f(x,s(y)) -> f(y,x)
r6: f(x,y) -> ack(x,y)
r7: ack(s(x),y) -> f(x,x)

The estimated dependency graph contains the following SCCs:

  {p2}


-- 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))
r4: f(s(x),y) -> f(x,s(x))
r5: f(x,s(y)) -> f(y,x)
r6: f(x,y) -> ack(x,y)
r7: ack(s(x),y) -> f(x,x)

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#_A(x1,x2) = x2
            s_A(x1) = max{12, x1 + 1}
            ack#!_A(x1,x2) = max{0, x1 - 15, x2 - 13}
            s!_A(x1) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ack#_A(x1,x2) = x2
            s_A(x1) = max{10, x1 + 4}
            ack#!_A(x1,x2) = max{1, x2 - 1}
            s!_A(x1) = max{2, x1 - 35}
        
  
    partial status:
  
      pi(ack#) = [1]
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.