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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ap#_A(x1,x2) = 0
              ff_1_A(x1) = 1
              ff_2#_A(x1,x2) = 0
              ap_A(x1,x2) = 1
              cons_2_A(x1,x2) = 0
              nil_0_A() = 1
              cons_0_A() = 1
              cons_1_A(x1) = 1
              ff_0_A() = 1
              ff_2_A(x1,x2) = 1
              ap#!_A(x1,x2) = x1 + x2 + 2
              ff_1!_A(x1) = 0
              ff_2#!_A(x1,x2) = x2 + 3
              ap!_A(x1,x2) = 2
              cons_2!_A(x1,x2) = 0
              nil_0!_A() = 0
              cons_0!_A() = 2
              cons_1!_A(x1) = 1
              ff_0!_A() = 1
              ff_2!_A(x1,x2) = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ap#_A(x1,x2) = max{21, x2 + 11}
            ff_1_A(x1) = max{10, x1 - 12}
            ff_2#_A(x1,x2) = max{21, x2 - 1}
            ap_A(x1,x2) = max{4, x1 - 15, x2 - 11}
            cons_2_A(x1,x2) = max{4, x1 - 32, x2 - 11}
            nil_0_A = 1
            cons_0_A = 1
            cons_1_A(x1) = max{0, x1 - 17}
            ff_0_A = 25
            ff_2_A(x1,x2) = max{4, x1 - 44, x2 - 22}
            ap#!_A(x1,x2) = max{x1 + 22, x2 + 22}
            ff_1!_A(x1) = 21
            ff_2#!_A(x1,x2) = max{32, x2 + 22}
            ap!_A(x1,x2) = max{18, x1 + 15}
            cons_2!_A(x1,x2) = max{19, x1 - 1, x2 - 2}
            nil_0!_A = 20
            cons_0!_A = 17
            cons_1!_A(x1) = 17
            ff_0!_A = 1
            ff_2!_A(x1,x2) = max{20, x1 + 2}
        
  
    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:

  p3



-- 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#(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}


-- 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))

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

  (no rules)

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ap#_A(x1,x2) = x2
              ff_1_A(x1) = x1
              ff_2#_A(x1,x2) = x2
              ap#!_A(x1,x2) = 0
              ff_1!_A(x1) = 3
              ff_2#!_A(x1,x2) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ap#_A(x1,x2) = max{1, x1 - 5, x2 - 3}
            ff_1_A(x1) = 4
            ff_2#_A(x1,x2) = max{1, x2 - 5}
            ap#!_A(x1,x2) = x2 + 3
            ff_1!_A(x1) = 15
            ff_2#!_A(x1,x2) = 8
        
  
    partial status:
  
      pi(ap#) = [1]
      pi(ff_1) = [1]
      pi(ff_2#) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.