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

  r1, r2, r3, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = x1 + 5
        fmap_1_A(x1) = max{3, x1 - 2}
        fmap_2#_A(x1,x2) = max{2, x1 - 3}
        fcons_2_A(x1,x2) = max{x1 + 9, x2 + 1}
        app_A(x1,x2) = max{12, x1 - 12, x2 - 1}
        cons_0_A = 10
        cons_1_A(x1) = max{11, x1 - 2}
        cons_2_A(x1,x2) = max{10, x1 - 26}
        fcons_0_A = 5
        fcons_1_A(x1) = max{6, x1 - 2}
        fmap_0_A = 2
  
    base order 2:
  
      precedence:
      
        app = cons_0 > cons_1 = fcons_0 > app# = fmap_1 = cons_2 > fcons_2 = fmap_0 > fmap_2# = fcons_1
      
  
    partial status:
  
      pi(app#) = [1]
      pi(fmap_1) = []
      pi(fmap_2#) = []
      pi(fcons_2) = []
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(fcons_0) = []
      pi(fcons_1) = []
      pi(fmap_0) = []

The following rules are removed:

  p2



-- 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)

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}


-- 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) -> 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

  r1, r2, r3, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{6, x1 - 2}
        fmap_1_A(x1) = max{10, x1 - 3}
        fmap_2#_A(x1,x2) = max{7, x1 - 6}
        fcons_2_A(x1,x2) = max{x1 + 5, x2 - 22}
        app_A(x1,x2) = max{18, x1 - 11, x2 - 2}
        cons_0_A = 10
        cons_1_A(x1) = max{18, x1 - 2}
        cons_2_A(x1,x2) = max{6, x1 - 41, x2 - 42}
        fcons_0_A = 9
        fcons_1_A(x1) = max{10, x1 - 3}
        fmap_0_A = 9
  
    base order 2:
  
      precedence:
      
        fcons_1 > app = fcons_0 > app# = fmap_1 = cons_2 > fmap_2# = fcons_2 = cons_1 = fmap_0 > cons_0
      
  
    partial status:
  
      pi(app#) = []
      pi(fmap_1) = []
      pi(fmap_2#) = []
      pi(fcons_2) = [1]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(fcons_0) = []
      pi(fcons_1) = []
      pi(fmap_0) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: app#(fmap_1(x1),y) -> fmap_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(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:

  (no SCCs)