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()))

Instead we prove termination of the uncurried system:

  ap(cons_0(),y) -> cons_1(y)
  ap(cons_1(x1),y) -> cons_2(x1,y)
  ap(ff_0(),y) -> ff_1(y)
  ap(ff_1(x1),y) -> ff_2(x1,y)
  ff_2(x,x) -> ap(ap(x,ff_1(x)),cons_2(x,nil_0()))


-- SCC decomposition.

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

p1: ap#(ff_1(x1),y) -> ff_2#(x1,y)
p2: ff_2#(x,x) -> ap#(ap(x,ff_1(x)),cons_2(x,nil_0()))
p3: ff_2#(x,x) -> ap#(x,ff_1(x))

and R consists of:

r1: ap(cons_0(),y) -> cons_1(y)
r2: ap(cons_1(x1),y) -> cons_2(x1,y)
r3: ap(ff_0(),y) -> ff_1(y)
r4: ap(ff_1(x1),y) -> ff_2(x1,y)
r5: ff_2(x,x) -> ap(ap(x,ff_1(x)),cons_2(x,nil_0()))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: ap#(ff_1(x1),y) -> ff_2#(x1,y)
p2: ff_2#(x,x) -> ap#(x,ff_1(x))
p3: ff_2#(x,x) -> ap#(ap(x,ff_1(x)),cons_2(x,nil_0()))

and R consists of:

r1: ap(cons_0(),y) -> cons_1(y)
r2: ap(cons_1(x1),y) -> cons_2(x1,y)
r3: ap(ff_0(),y) -> ff_1(y)
r4: ap(ff_1(x1),y) -> ff_2(x1,y)
r5: ff_2(x,x) -> ap(ap(x,ff_1(x)),cons_2(x,nil_0()))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ap#_A(x1,x2) = max{x1 + 47, x2 + 47}
        ff_1_A(x1) = max{13, x1 - 18}
        ff_2#_A(x1,x2) = max{60, x1 + 14, x2 + 47}
        ap_A(x1,x2) = max{3, x1 - 8, x2 - 14}
        cons_2_A(x1,x2) = max{1, x2 - 49}
        nil_0_A = 1
        cons_0_A = 11
        cons_1_A(x1) = max{2, x1 - 15}
        ff_0_A = 22
        ff_2_A(x1,x2) = max{4, x2 - 15}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          ap#_A(x1,x2) = 1
          ff_1_A(x1) = max{11, x1 + 9}
          ff_2#_A(x1,x2) = 1
          ap_A(x1,x2) = max{23, x2 + 10}
          cons_2_A(x1,x2) = max{13, x2 + 10}
          nil_0_A = 2
          cons_0_A = 23
          cons_1_A(x1) = max{23, x1 + 10}
          ff_0_A = 4
          ff_2_A(x1,x2) = 23
          ap#!_A(x1,x2) = max{1, x1 - 4}
          ff_1!_A(x1) = x1 + 6
          ff_2#!_A(x1,x2) = max{6, x1 - 3}
          ap!_A(x1,x2) = max{3, x1 - 5}
          cons_2!_A(x1,x2) = max{4, x1 - 7}
          nil_0!_A = 1
          cons_0!_A = 1
          cons_1!_A(x1) = max{18, x1 - 8}
          ff_0!_A = 1
          ff_2!_A(x1,x2) = max{7, x2 - 6}
  
    partial status:
  
      pi(ap#) = []
      pi(ff_1) = []
      pi(ff_2#) = []
      pi(ap) = []
      pi(cons_2) = []
      pi(nil_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(ff_0) = []
      pi(ff_2) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.