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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ap#_A(x1,x2) = max{1, x1 - 1, x2 - 1}
          ap_A(x1,x2) = max{4, x1 - 5, x2 - 1}
          ff_A = 18
          cons_A = 7
          nil_A = 0
    
      precedence:
      
        nil > ap# = ap = ff = cons
      
      partial status:
    
        pi(ap#) = []
        pi(ap) = []
        pi(ff) = []
        pi(cons) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ap#_A(x1,x2) = 0
          ap_A(x1,x2) = 14
          ff_A = 26
          cons_A = 38
          nil_A = 35
    
      precedence:
      
        cons = nil > ff > ap > ap#
      
      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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ap#_A(x1,x2) = max{x1 + 8, x2 + 7}
          ap_A(x1,x2) = max{x1 + 3, x2 + 3}
          ff_A = 2
    
      precedence:
      
        ap# = ap = ff
      
      partial status:
    
        pi(ap#) = [1, 2]
        pi(ap) = [1, 2]
        pi(ff) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ap#_A(x1,x2) = max{x1 + 4, x2 + 7}
          ap_A(x1,x2) = max{x1 + 1, x2 + 3}
          ff_A = 2
    
      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.