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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{0, x1 - 14, x2 - 4}
        app_A(x1,x2) = max{x1 + 8, x2 + 13}
        mapt_A = 11
        leaf_A = 0
        maptlist_A = 4
        cons_A = 12
        node_A = 5
  
    precedence:
    
      app# = app = mapt = leaf = maptlist = cons = node
    
    partial status:
  
      pi(app#) = []
      pi(app) = [1, 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{5, x1 - 2, x2 + 2}
        app_A(x1,x2) = max{10, x1 + 6, x2 + 9}
        maptlist_A = 0
        cons_A = 4
        mapt_A = 14
        node_A = 3
  
    precedence:
    
      app# = cons = mapt > app = node > maptlist
    
    partial status:
  
      pi(app#) = [2]
      pi(app) = []
      pi(maptlist) = []
      pi(cons) = []
      pi(mapt) = []
      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(mapt(),f),x)
p2: 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}


-- 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(mapt(),f),x)
p2: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{40, x1 + 12, x2 + 22}
        app_A(x1,x2) = max{x1 + 6, x2}
        maptlist_A = 3
        cons_A = 6
        mapt_A = 5
        node_A = 4
  
    precedence:
    
      app# = app = maptlist = cons = mapt = node
    
    partial status:
  
      pi(app#) = [2]
      pi(app) = [1, 2]
      pi(maptlist) = []
      pi(cons) = []
      pi(mapt) = []
      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(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:

  (no SCCs)