YES

We show the termination of the TRS R:

  ap(ap(ff(),x),x) -> ap(ap(x,ap(ff(),x)),ap(ap(cons(),x),nil()))

-- SCC decomposition.

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

p1: ap#(ap(ff(),x),x) -> ap#(ap(x,ap(ff(),x)),ap(ap(cons(),x),nil()))
p2: ap#(ap(ff(),x),x) -> ap#(x,ap(ff(),x))
p3: ap#(ap(ff(),x),x) -> ap#(ap(cons(),x),nil())
p4: ap#(ap(ff(),x),x) -> ap#(cons(),x)

and R consists of:

r1: ap(ap(ff(),x),x) -> ap(ap(x,ap(ff(),x)),ap(ap(cons(),x),nil()))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: ap#(ap(ff(),x),x) -> ap#(ap(x,ap(ff(),x)),ap(ap(cons(),x),nil()))
p2: ap#(ap(ff(),x),x) -> ap#(x,ap(ff(),x))

and R consists of:

r1: ap(ap(ff(),x),x) -> ap(ap(x,ap(ff(),x)),ap(ap(cons(),x),nil()))

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      ap#_A(x1,x2) = max{0, x1 - 5, x2 - 5}
      ap_A(x1,x2) = max{4, x1 - 3, x2 - 5}
      ff_A = 14
      cons_A = 0
      nil_A = 8
    
    2. max/plus interpretations on natural numbers:
    
      ap#_A(x1,x2) = 0
      ap_A(x1,x2) = 0
      ff_A = 0
      cons_A = 0
      nil_A = 0
    

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: ap#(ap(ff(),x),x) -> ap#(x,ap(ff(),x))

and R consists of:

r1: ap(ap(ff(),x),x) -> ap(ap(x,ap(ff(),x)),ap(ap(cons(),x),nil()))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: ap#(ap(ff(),x),x) -> ap#(x,ap(ff(),x))

and R consists of:

r1: ap(ap(ff(),x),x) -> ap(ap(x,ap(ff(),x)),ap(ap(cons(),x),nil()))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      ap#_A(x1,x2) = max{0, x1 - 2}
      ap_A(x1,x2) = max{x1 + 3, x2 + 5}
      ff_A = 0
    
    2. max/plus interpretations on natural numbers:
    
      ap#_A(x1,x2) = 0
      ap_A(x1,x2) = 0
      ff_A = 0
    

The next rules are strictly ordered:

  p1

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