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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      ap#_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,1,0)) x1 + ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0)) x2
      ap_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(1,0,0,0)) x1 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (0,0,1,1)
      ff_A() = (3,1,1,1)
      cons_A() = (1,1,1,0)
      nil_A() = (1,1,1,1)

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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      ap#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,1,1,1)) x2
      ap_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)
      ff_A() = (1,1,1,1)

The next rules are strictly ordered:

  p1

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