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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        ap#_A(x1,x2) = max{1, x1 - 27, x2 - 27}
        ap_A(x1,x2) = max{16, x1 - 5, x2 - 10}
        ff_A = 35
        cons_A = 0
        nil_A = 2
  
    precedence:
    
      ap# = ap = ff > cons = nil
    
    partial status:
  
      pi(ap#) = []
      pi(ap) = []
      pi(ff) = []
      pi(cons) = []
      pi(nil) = []

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        ap#_A(x1,x2) = max{16, x1 + 7, x2 + 2}
        ap_A(x1,x2) = max{15, x1 + 10, x2 + 7}
        ff_A = 1
  
    precedence:
    
      ap# = ap = ff
    
    partial status:
  
      pi(ap#) = []
      pi(ap) = [2]
      pi(ff) = []

The next rules are strictly ordered:

  p1

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