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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          ap#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (4,1)
          ap_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(0,1)) x2
          ff_A() = (3,2)
          cons_A() = (8,4)
          nil_A() = (0,0)
  
    precedence:
    
      ff = nil > cons > ap# = ap
    
    partial status:
  
      pi(ap#) = []
      pi(ap) = []
      pi(ff) = []
      pi(cons) = []
      pi(nil) = []

The next rules are strictly ordered:

  p2

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#(ap(x,ap(ff(),x)),ap(ap(cons(),x),nil()))

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#(ap(x,ap(ff(),x)),ap(ap(cons(),x),nil()))

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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          ap#_A(x1,x2) = x1 + ((0,1),(0,0)) x2 + (2,2)
          ap_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,0),(0,1)) x2 + (3,0)
          ff_A() = (1,1)
          cons_A() = (7,1)
          nil_A() = (2,0)
  
    precedence:
    
      ap# = ff > ap = nil > cons
    
    partial status:
  
      pi(ap#) = [1]
      pi(ap) = []
      pi(ff) = []
      pi(cons) = []
      pi(nil) = []

The next rules are strictly ordered:

  p1

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