YES
We show the termination of the TRS R:

  app(app(fmap(),fnil()),x) -> nil()
  app(app(fmap(),app(app(fcons(),f),t)),x) -> app(app(cons(),app(f,x)),app(app(fmap(),t),x))

Instead we prove termination of the uncurried system:

  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(fcons_0(),y) -> fcons_1(y)
  app(fcons_1(x1),y) -> fcons_2(x1,y)
  app(fmap_0(),y) -> fmap_1(y)
  app(fmap_1(x1),y) -> fmap_2(x1,y)
  fmap_2(fnil_0(),x) -> nil_0()
  fmap_2(fcons_2(f,t),x) -> cons_2(app(f,x),fmap_2(t,x))


-- SCC decomposition.

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

p1: app#(fmap_1(x1),y) -> fmap_2#(x1,y)
p2: fmap_2#(fcons_2(f,t),x) -> app#(f,x)
p3: fmap_2#(fcons_2(f,t),x) -> fmap_2#(t,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(fcons_0(),y) -> fcons_1(y)
r4: app(fcons_1(x1),y) -> fcons_2(x1,y)
r5: app(fmap_0(),y) -> fmap_1(y)
r6: app(fmap_1(x1),y) -> fmap_2(x1,y)
r7: fmap_2(fnil_0(),x) -> nil_0()
r8: fmap_2(fcons_2(f,t),x) -> cons_2(app(f,x),fmap_2(t,x))

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: app#(fmap_1(x1),y) -> fmap_2#(x1,y)
p2: fmap_2#(fcons_2(f,t),x) -> fmap_2#(t,x)
p3: fmap_2#(fcons_2(f,t),x) -> app#(f,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(fcons_0(),y) -> fcons_1(y)
r4: app(fcons_1(x1),y) -> fcons_2(x1,y)
r5: app(fmap_0(),y) -> fmap_1(y)
r6: app(fmap_1(x1),y) -> fmap_2(x1,y)
r7: fmap_2(fnil_0(),x) -> nil_0()
r8: fmap_2(fcons_2(f,t),x) -> cons_2(app(f,x),fmap_2(t,x))

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!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 8, x2 - 1}
            fmap_1_A(x1) = x1 + 3
            fmap_2#_A(x1,x2) = max{0, x1 - 5, x2 - 1}
            fcons_2_A(x1,x2) = max{4, x1 - 3, x2}
            app#!_A(x1,x2) = max{0, x1 - 5}
            fmap_1!_A(x1) = max{2, x1 - 10}
            fmap_2#!_A(x1,x2) = 1
            fcons_2!_A(x1,x2) = max{0, x1 - 4, x2 - 10}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 5, x2 - 1}
            fmap_1_A(x1) = x1 + 3
            fmap_2#_A(x1,x2) = max{0, x1 - 2, x2 - 1}
            fcons_2_A(x1,x2) = max{1, x1 - 3, x2}
            app#!_A(x1,x2) = max{1, x1 - 7}
            fmap_1!_A(x1) = max{3, x1 - 5}
            fmap_2#!_A(x1,x2) = max{0, x1 - 5, x2 - 1}
            fcons_2!_A(x1,x2) = max{2, x1 - 6, x2 - 10}
        
  
    partial status:
  
      pi(app#) = [1]
      pi(fmap_1) = [1]
      pi(fmap_2#) = [1]
      pi(fcons_2) = [1, 2]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.