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 combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (14,5)
            app_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (6,2)
            mapt_A() = (7,3)
            leaf_A() = (1,1)
            maptlist_A() = (0,0)
            cons_A() = (22,18)
            node_A() = (5,1)
    
      precedence:
      
        mapt = leaf = cons = node > app# = app = maptlist
      
      partial status:
    
        pi(app#) = [2]
        pi(app) = []
        pi(mapt) = []
        pi(leaf) = []
        pi(maptlist) = []
        pi(cons) = []
        pi(node) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (4,4)
            app_A(x1,x2) = (1,1)
            mapt_A() = (5,5)
            leaf_A() = (5,5)
            maptlist_A() = (2,2)
            cons_A() = (6,6)
            node_A() = (3,3)
    
      precedence:
      
        app > cons > leaf > app# > node > mapt > maptlist
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(mapt) = []
        pi(leaf) = []
        pi(maptlist) = []
        pi(cons) = []
        pi(node) = []
    

The next rules are strictly ordered:

  p1, p2, p4

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(mapt(),f),x)

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:

  (no SCCs)