YES
We show the termination of the TRS R:

  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))

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(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))


-- SCC decomposition.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)
p2: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: map_2(f,nil_0()) -> nil_0()
r6: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))

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#(map_1(x1),y) -> map_2#(x1,y)
p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p3: map_2#(f,cons_2(x,xs)) -> 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(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: map_2(f,nil_0()) -> nil_0()
r6: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))

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 - 3, x2 - 1}
            map_1_A(x1) = x1 + 2
            map_2#_A(x1,x2) = max{0, x1 - 3, x2 - 5}
            cons_2_A(x1,x2) = max{x1 + 4, x2 + 1}
            app#!_A(x1,x2) = max{5, x1 - 3, x2 - 10}
            map_1!_A(x1) = max{4, x1 - 15}
            map_2#!_A(x1,x2) = max{5, x1 - 2, x2 - 14}
            cons_2!_A(x1,x2) = max{2, x1 - 9, x2 - 7}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 4, x2 - 2}
            map_1_A(x1) = max{3, x1 + 2}
            map_2#_A(x1,x2) = max{0, x1 - 4, x2 - 6}
            cons_2_A(x1,x2) = max{5, x1 + 4, x2 + 2}
            app#!_A(x1,x2) = max{7, x1 - 5, x2 - 2}
            map_1!_A(x1) = max{2, x1 - 4}
            map_2#!_A(x1,x2) = max{7, x1 - 5, x2 - 6}
            cons_2!_A(x1,x2) = max{x1 + 6, x2 + 4}
        
  
    partial status:
  
      pi(app#) = [2]
      pi(map_1) = [1]
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]

The following rules are removed:

  p2, p3



-- SCC decomposition.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: map_2(f,nil_0()) -> nil_0()
r6: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))

The estimated dependency graph contains the following SCCs:

  (no SCCs)