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

  r1, r2, r3, r4, r5, r6

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            app#_A(0,0) = 0
            app#_A(0,1) = 0
            app#_A(1,0) = 0
            app#_A(1,1) = 0
            map_1_A(0) = 1
            map_1_A(1) = 1
            map_2#_A(0,0) = 0
            map_2#_A(0,1) = 0
            map_2#_A(1,0) = 0
            map_2#_A(1,1) = 0
            cons_2_A(0,0) = 0
            cons_2_A(0,1) = 0
            cons_2_A(1,0) = 0
            cons_2_A(1,1) = 0
            app_A(0,0) = 1
            app_A(0,1) = 1
            app_A(1,0) = 0
            app_A(1,1) = 0
            cons_0_A = 0
            cons_1_A(0) = 1
            cons_1_A(1) = 1
            map_0_A = 0
            map_2_A(0,0) = 0
            map_2_A(0,1) = 0
            map_2_A(1,0) = 0
            map_2_A(1,1) = 0
            nil_0_A = 0
            app#!_A(0,0) = 0
            app#!_A(0,1) = 1
            app#!_A(1,0) = 1
            app#!_A(1,1) = 1
            map_1!_A(0) = 0
            map_1!_A(1) = 0
            map_2#!_A(0,0) = 1
            map_2#!_A(0,1) = 1
            map_2#!_A(1,0) = 1
            map_2#!_A(1,1) = 0
            cons_2!_A(0,0) = 0
            cons_2!_A(0,1) = 1
            cons_2!_A(1,0) = 0
            cons_2!_A(1,1) = 1
            app!_A(0,0) = 1
            app!_A(0,1) = 1
            app!_A(1,0) = 1
            app!_A(1,1) = 1
            cons_0!_A = 1
            cons_1!_A(0) = 1
            cons_1!_A(1) = 1
            map_0!_A = 1
            map_2!_A(0,0) = 0
            map_2!_A(0,1) = 0
            map_2!_A(1,0) = 0
            map_2!_A(1,1) = 0
            nil_0!_A = 0
  
    partial status:
  
      pi(app#) = [1]
      pi(map_1) = [1]
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app) = [1, 2]
      pi(cons_0) = []
      pi(cons_1) = []
      pi(map_0) = []
      pi(map_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p2: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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 set of usable rules consists of

  r1, r2, r3, r4, r5, r6

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            map_2#_A(0,0) = 0
            map_2#_A(0,1) = 0
            map_2#_A(1,0) = 0
            map_2#_A(1,1) = 0
            cons_2_A(0,0) = 0
            cons_2_A(0,1) = 1
            cons_2_A(1,0) = 0
            cons_2_A(1,1) = 1
            app_A(0,0) = 0
            app_A(0,1) = 1
            app_A(1,0) = 0
            app_A(1,1) = 0
            cons_0_A = 1
            cons_1_A(0) = 0
            cons_1_A(1) = 0
            map_0_A = 1
            map_1_A(0) = 0
            map_1_A(1) = 0
            map_2_A(0,0) = 0
            map_2_A(0,1) = 1
            map_2_A(1,0) = 0
            map_2_A(1,1) = 1
            nil_0_A = 0
            map_2#!_A(0,0) = 1
            map_2#!_A(0,1) = 0
            map_2#!_A(1,0) = 1
            map_2#!_A(1,1) = 0
            cons_2!_A(0,0) = 1
            cons_2!_A(0,1) = 0
            cons_2!_A(1,0) = 1
            cons_2!_A(1,1) = 0
            app!_A(0,0) = 1
            app!_A(0,1) = 1
            app!_A(1,0) = 0
            app!_A(1,1) = 1
            cons_0!_A = 1
            cons_1!_A(0) = 0
            cons_1!_A(1) = 0
            map_0!_A = 1
            map_1!_A(0) = 0
            map_1!_A(1) = 0
            map_2!_A(0,0) = 1
            map_2!_A(0,1) = 0
            map_2!_A(1,0) = 1
            map_2!_A(1,1) = 0
            nil_0!_A = 1
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = [2]
      pi(cons_0) = []
      pi(cons_1) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(map_2) = [2]
      pi(nil_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.