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))

-- SCC decomposition.

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

p1: app#(app(fmap(),app(app(fcons(),f),t)),x) -> app#(app(cons(),app(f,x)),app(app(fmap(),t),x))
p2: app#(app(fmap(),app(app(fcons(),f),t)),x) -> app#(cons(),app(f,x))
p3: app#(app(fmap(),app(app(fcons(),f),t)),x) -> app#(f,x)
p4: app#(app(fmap(),app(app(fcons(),f),t)),x) -> app#(app(fmap(),t),x)
p5: app#(app(fmap(),app(app(fcons(),f),t)),x) -> app#(fmap(),t)

and R consists of:

r1: app(app(fmap(),fnil()),x) -> nil()
r2: app(app(fmap(),app(app(fcons(),f),t)),x) -> app(app(cons(),app(f,x)),app(app(fmap(),t),x))

The estimated dependency graph contains the following SCCs:

  {p3, p4}


-- Reduction pair.

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

p1: app#(app(fmap(),app(app(fcons(),f),t)),x) -> app#(f,x)
p2: app#(app(fmap(),app(app(fcons(),f),t)),x) -> app#(app(fmap(),t),x)

and R consists of:

r1: app(app(fmap(),fnil()),x) -> nil()
r2: app(app(fmap(),app(app(fcons(),f),t)),x) -> app(app(cons(),app(f,x)),app(app(fmap(),t),x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = x1
      app_A(x1,x2) = x1 + x2 + (0,1)
      fmap_A() = (1,1)
      fcons_A() = (1,0)

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.