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 - 7}
            s_A(x1) = x1 + 6
            |0|_A = 0
            f#_A(x1,x2) = max{1, x1 - 7, x2 - 7}
            ack_A(x1,x2) = max{7, x1, x2 + 6}
            ack#!_A(x1,x2) = max{2, x1 - 2}
            s!_A(x1) = max{9, x1 + 8}
            |0|!_A = 3
            f#!_A(x1,x2) = max{3, x1 + 1, x2 + 1}
            ack!_A(x1,x2) = max{x1 - 1, x2 + 10}
        
        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) = 0
              s_A(x1) = 1
              |0|_A() = 1
              f#_A(x1,x2) = 0
              ack_A(x1,x2) = 1
              ack#!_A(x1,x2) = 1
              s!_A(x1) = 3
              |0|!_A() = 0
              f#!_A(x1,x2) = 2
              ack!_A(x1,x2) = x1 + 4
        
  
    partial status:
  
      pi(ack#) = []
      pi(s) = [1]
      pi(|0|) = []
      pi(f#) = []
      pi(ack) = [1, 2]

The following rules are removed:

  p1, p2, p3, p5



-- SCC decomposition.

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

p1: ack#(s(x),s(y)) -> ack#(s(x),y)
p2: f#(x,s(y)) -> f#(y,x)
p3: 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 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),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) = 0
            s_A(x1) = x1 + 7
            ack#!_A(x1,x2) = max{1, x2 - 8}
            s!_A(x1) = x1 + 11
        
        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 + 3
              ack#!_A(x1,x2) = 3
              s!_A(x1) = 2
        
  
    partial status:
  
      pi(ack#) = [2]
      pi(s) = [1]

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: f#(x,s(y)) -> f#(y,x)
p2: 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

  (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:
          
            f#_A(x1,x2) = max{0, x1 - 7, x2 - 7}
            s_A(x1) = max{6, x1 + 2}
            f#!_A(x1,x2) = max{x1 + 3, x2 + 2}
            s!_A(x1) = 10
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1,x2) = 0
              s_A(x1) = x1 + 2
              f#!_A(x1,x2) = 2
              s!_A(x1) = 1
        
  
    partial status:
  
      pi(f#) = []
      pi(s) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.