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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ap#_A(x1,x2) = max{0, x1 - 9, x2 - 9}
            ff_1_A(x1) = max{8, x1 - 10}
            ff_2#_A(x1,x2) = max{0, x1 - 19, x2 - 9}
            ap_A(x1,x2) = max{6, x1 - 22, x2 - 10}
            cons_2_A(x1,x2) = max{6, x1 - 35}
            nil_0_A = 0
            cons_0_A = 0
            cons_1_A(x1) = max{6, x1 - 13}
            ff_0_A = 30
            ff_2_A(x1,x2) = max{6, x1 - 42, x2 - 45}
            ap#!_A(x1,x2) = max{x1 + 21, x2 + 21}
            ff_1!_A(x1) = 0
            ff_2#!_A(x1,x2) = max{29, x1 + 2, x2 + 21}
            ap!_A(x1,x2) = x1 + 19
            cons_2!_A(x1,x2) = max{7, x1 + 3}
            nil_0!_A = 28
            cons_0!_A = 2
            cons_1!_A(x1) = 1
            ff_0!_A = 0
            ff_2!_A(x1,x2) = max{26, x1 + 8}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ap#_A(x1,x2) = 1
            ff_1_A(x1) = max{16, x1 + 3}
            ff_2#_A(x1,x2) = 1
            ap_A(x1,x2) = max{x1 - 17, x2 + 3}
            cons_2_A(x1,x2) = max{0, x1 - 17, x2 - 4}
            nil_0_A = 0
            cons_0_A = 33
            cons_1_A(x1) = max{16, x1 + 3}
            ff_0_A = 33
            ff_2_A(x1,x2) = max{3, x1 - 14, x2 - 2}
            ap#!_A(x1,x2) = max{1, x1 - 2, x2 - 21}
            ff_1!_A(x1) = 11
            ff_2#!_A(x1,x2) = max{13, x1 - 1}
            ap!_A(x1,x2) = max{x1 - 2, x2 + 10}
            cons_2!_A(x1,x2) = max{13, x1 - 1}
            nil_0!_A = 13
            cons_0!_A = 1
            cons_1!_A(x1) = x1 + 10
            ff_0!_A = 11
            ff_2!_A(x1,x2) = max{12, x2 + 5}
        
  
    partial status:
  
      pi(ap#) = [2]
      pi(ff_1) = []
      pi(ff_2#) = [2]
      pi(ap) = [2]
      pi(cons_2) = []
      pi(nil_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(ff_0) = []
      pi(ff_2) = [2]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.