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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ap#_A(x1,x2) = max{x1 + 30, x2 + 30}
        ff_1_A(x1) = max{12, x1 - 16}
        ff_2#_A(x1,x2) = max{42, x1 + 13, x2 + 30}
        ap_A(x1,x2) = max{1, x1 - 9, x2 - 15}
        cons_2_A(x1,x2) = max{1, x1 - 31, x2 - 30}
        nil_0_A = 31
        cons_0_A = 8
        cons_1_A(x1) = max{0, x1 - 22}
        ff_0_A = 22
        ff_2_A(x1,x2) = max{2, x1 - 26, x2 - 17}
  
    base order 2:
  
      precedence:
      
        ap# = ff_1 = ff_2# > nil_0 > ap = cons_0 = ff_0 = ff_2 > cons_2 > cons_1
      
  
    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

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        ap#_A(x1,x2) = max{2, x1 - 11, x2 - 16}
        ff_1_A(x1) = x1 + 15
        ff_2#_A(x1,x2) = x1 + 3
        ap_A(x1,x2) = max{9, x1 - 16, x2 - 24}
        cons_0_A = 15
        cons_1_A(x1) = max{8, x1 - 25}
        cons_2_A(x1,x2) = max{7, x1 - 42}
  
    base order 2:
  
      precedence:
      
        ff_2# > ap# = ap > ff_1 = cons_1 > cons_0 = cons_2
      
  
    partial status:
  
      pi(ap#) = []
      pi(ff_1) = []
      pi(ff_2#) = [1]
      pi(ap) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: ap#(ff_1(x1),y) -> ff_2#(x1,y)

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:

  (no SCCs)