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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{0, x1 - 13}
        app_A(x1,x2) = max{x1 + 4, x2 + 6}
        fmap_A = 2
        fcons_A = 3
  
    precedence:
    
      app# = app = fmap = fcons
    
    partial status:
  
      pi(app#) = []
      pi(app) = [1, 2]
      pi(fmap) = []
      pi(fcons) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: app#(app(fmap(),app(app(fcons(),f),t)),x) -> app#(f,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 estimated dependency graph contains the following SCCs:

  {p1}


-- 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)

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{0, x1 - 2}
        app_A(x1,x2) = max{4, x1 - 3, x2 + 2}
        fmap_A = 6
        fcons_A = 7
  
    precedence:
    
      app# = app = fmap = fcons
    
    partial status:
  
      pi(app#) = []
      pi(app) = [2]
      pi(fmap) = []
      pi(fcons) = []

The next rules are strictly ordered:

  p1

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