YES

We show the termination of the TRS R:

  app(app(mapt(),f),app(leaf(),x)) -> app(leaf(),app(f,x))
  app(app(mapt(),f),app(node(),xs)) -> app(node(),app(app(maptlist(),f),xs))
  app(app(maptlist(),f),nil()) -> nil()
  app(app(maptlist(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(app(mapt(),f),x)),app(app(maptlist(),f),xs))

-- SCC decomposition.

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

p1: app#(app(mapt(),f),app(leaf(),x)) -> app#(leaf(),app(f,x))
p2: app#(app(mapt(),f),app(leaf(),x)) -> app#(f,x)
p3: app#(app(mapt(),f),app(node(),xs)) -> app#(node(),app(app(maptlist(),f),xs))
p4: app#(app(mapt(),f),app(node(),xs)) -> app#(app(maptlist(),f),xs)
p5: app#(app(mapt(),f),app(node(),xs)) -> app#(maptlist(),f)
p6: app#(app(maptlist(),f),app(app(cons(),x),xs)) -> app#(app(cons(),app(app(mapt(),f),x)),app(app(maptlist(),f),xs))
p7: app#(app(maptlist(),f),app(app(cons(),x),xs)) -> app#(cons(),app(app(mapt(),f),x))
p8: app#(app(maptlist(),f),app(app(cons(),x),xs)) -> app#(app(mapt(),f),x)
p9: app#(app(maptlist(),f),app(app(cons(),x),xs)) -> app#(mapt(),f)
p10: app#(app(maptlist(),f),app(app(cons(),x),xs)) -> app#(app(maptlist(),f),xs)

and R consists of:

r1: app(app(mapt(),f),app(leaf(),x)) -> app(leaf(),app(f,x))
r2: app(app(mapt(),f),app(node(),xs)) -> app(node(),app(app(maptlist(),f),xs))
r3: app(app(maptlist(),f),nil()) -> nil()
r4: app(app(maptlist(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(app(mapt(),f),x)),app(app(maptlist(),f),xs))

The estimated dependency graph contains the following SCCs:

  {p2, p4, p8, p10}


-- Reduction pair.

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

p1: app#(app(mapt(),f),app(leaf(),x)) -> app#(f,x)
p2: app#(app(maptlist(),f),app(app(cons(),x),xs)) -> app#(app(maptlist(),f),xs)
p3: app#(app(maptlist(),f),app(app(cons(),x),xs)) -> app#(app(mapt(),f),x)
p4: app#(app(mapt(),f),app(node(),xs)) -> app#(app(maptlist(),f),xs)

and R consists of:

r1: app(app(mapt(),f),app(leaf(),x)) -> app(leaf(),app(f,x))
r2: app(app(mapt(),f),app(node(),xs)) -> app(node(),app(app(maptlist(),f),xs))
r3: app(app(maptlist(),f),nil()) -> nil()
r4: app(app(maptlist(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(app(mapt(),f),x)),app(app(maptlist(),f),xs))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      app > app# > cons > maptlist > leaf > node > mapt
    
    argument filter:
  
      pi(app#) = [1]
      pi(app) = [2]
      pi(mapt) = []
      pi(leaf) = []
      pi(maptlist) = []
      pi(cons) = []
      pi(node) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: app#(app(maptlist(),f),app(app(cons(),x),xs)) -> app#(app(maptlist(),f),xs)
p2: app#(app(maptlist(),f),app(app(cons(),x),xs)) -> app#(app(mapt(),f),x)
p3: app#(app(mapt(),f),app(node(),xs)) -> app#(app(maptlist(),f),xs)

and R consists of:

r1: app(app(mapt(),f),app(leaf(),x)) -> app(leaf(),app(f,x))
r2: app(app(mapt(),f),app(node(),xs)) -> app(node(),app(app(maptlist(),f),xs))
r3: app(app(maptlist(),f),nil()) -> nil()
r4: app(app(maptlist(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(app(mapt(),f),x)),app(app(maptlist(),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#(app(maptlist(),f),app(app(cons(),x),xs)) -> app#(app(maptlist(),f),xs)
p2: app#(app(maptlist(),f),app(app(cons(),x),xs)) -> app#(app(mapt(),f),x)
p3: app#(app(mapt(),f),app(node(),xs)) -> app#(app(maptlist(),f),xs)

and R consists of:

r1: app(app(mapt(),f),app(leaf(),x)) -> app(leaf(),app(f,x))
r2: app(app(mapt(),f),app(node(),xs)) -> app(node(),app(app(maptlist(),f),xs))
r3: app(app(maptlist(),f),nil()) -> nil()
r4: app(app(maptlist(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(app(mapt(),f),x)),app(app(maptlist(),f),xs))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      app > maptlist > app# > node > mapt > cons
    
    argument filter:
  
      pi(app#) = 2
      pi(app) = [1, 2]
      pi(maptlist) = []
      pi(cons) = []
      pi(mapt) = []
      pi(node) = []

The next rules are strictly ordered:

  p1, p2, p3

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