YES

We show the termination of the TRS R:

  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
  app(app(le(),|0|()),y) -> true()
  app(app(le(),app(s(),x)),|0|()) -> false()
  app(app(le(),app(s(),x)),app(s(),y)) -> app(app(le(),x),y)
  app(app(maxlist(),x),app(app(cons(),y),ys)) -> app(app(if(),app(app(le(),x),y)),app(app(maxlist(),y),ys))
  app(app(maxlist(),x),nil()) -> x
  app(height(),app(app(node(),x),xs)) -> app(s(),app(app(maxlist(),|0|()),app(app(map(),height()),xs)))

-- SCC decomposition.

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

p1: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(cons(),app(f,x)),app(app(map(),f),xs))
p2: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(cons(),app(f,x))
p3: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p4: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)
p5: app#(app(le(),app(s(),x)),app(s(),y)) -> app#(app(le(),x),y)
p6: app#(app(le(),app(s(),x)),app(s(),y)) -> app#(le(),x)
p7: app#(app(maxlist(),x),app(app(cons(),y),ys)) -> app#(app(if(),app(app(le(),x),y)),app(app(maxlist(),y),ys))
p8: app#(app(maxlist(),x),app(app(cons(),y),ys)) -> app#(if(),app(app(le(),x),y))
p9: app#(app(maxlist(),x),app(app(cons(),y),ys)) -> app#(app(le(),x),y)
p10: app#(app(maxlist(),x),app(app(cons(),y),ys)) -> app#(le(),x)
p11: app#(app(maxlist(),x),app(app(cons(),y),ys)) -> app#(app(maxlist(),y),ys)
p12: app#(app(maxlist(),x),app(app(cons(),y),ys)) -> app#(maxlist(),y)
p13: app#(height(),app(app(node(),x),xs)) -> app#(s(),app(app(maxlist(),|0|()),app(app(map(),height()),xs)))
p14: app#(height(),app(app(node(),x),xs)) -> app#(app(maxlist(),|0|()),app(app(map(),height()),xs))
p15: app#(height(),app(app(node(),x),xs)) -> app#(maxlist(),|0|())
p16: app#(height(),app(app(node(),x),xs)) -> app#(app(map(),height()),xs)
p17: app#(height(),app(app(node(),x),xs)) -> app#(map(),height())

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(le(),|0|()),y) -> true()
r4: app(app(le(),app(s(),x)),|0|()) -> false()
r5: app(app(le(),app(s(),x)),app(s(),y)) -> app(app(le(),x),y)
r6: app(app(maxlist(),x),app(app(cons(),y),ys)) -> app(app(if(),app(app(le(),x),y)),app(app(maxlist(),y),ys))
r7: app(app(maxlist(),x),nil()) -> x
r8: app(height(),app(app(node(),x),xs)) -> app(s(),app(app(maxlist(),|0|()),app(app(map(),height()),xs)))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p16}
  {p11}
  {p5}


-- Reduction pair.

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

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

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(le(),|0|()),y) -> true()
r4: app(app(le(),app(s(),x)),|0|()) -> false()
r5: app(app(le(),app(s(),x)),app(s(),y)) -> app(app(le(),x),y)
r6: app(app(maxlist(),x),app(app(cons(),y),ys)) -> app(app(if(),app(app(le(),x),y)),app(app(maxlist(),y),ys))
r7: app(app(maxlist(),x),nil()) -> x
r8: app(height(),app(app(node(),x),xs)) -> app(s(),app(app(maxlist(),|0|()),app(app(map(),height()),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 + x2 + (2,3)
            app_A(x1,x2) = x1 + x2 + (0,2)
            map_A() = (0,2)
            cons_A() = (1,8)
            height_A() = (1,1)
            node_A() = (0,9)
    
      precedence:
      
        app# = app = map = cons > height = node
      
      partial status:
    
        pi(app#) = [2]
        pi(app) = [1, 2]
        pi(map) = []
        pi(cons) = []
        pi(height) = []
        pi(node) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (5,5)
            app_A(x1,x2) = x2 + (2,2)
            map_A() = (5,5)
            cons_A() = (6,6)
            height_A() = (1,1)
            node_A() = (4,4)
    
      precedence:
      
        cons > app# = app = map = height > node
      
      partial status:
    
        pi(app#) = []
        pi(app) = [2]
        pi(map) = []
        pi(cons) = []
        pi(height) = []
        pi(node) = []
    

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p2: app#(height(),app(app(node(),x),xs)) -> app#(app(map(),height()),xs)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(le(),|0|()),y) -> true()
r4: app(app(le(),app(s(),x)),|0|()) -> false()
r5: app(app(le(),app(s(),x)),app(s(),y)) -> app(app(le(),x),y)
r6: app(app(maxlist(),x),app(app(cons(),y),ys)) -> app(app(if(),app(app(le(),x),y)),app(app(maxlist(),y),ys))
r7: app(app(maxlist(),x),nil()) -> x
r8: app(height(),app(app(node(),x),xs)) -> app(s(),app(app(maxlist(),|0|()),app(app(map(),height()),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(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p2: app#(height(),app(app(node(),x),xs)) -> app#(app(map(),height()),xs)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(le(),|0|()),y) -> true()
r4: app(app(le(),app(s(),x)),|0|()) -> false()
r5: app(app(le(),app(s(),x)),app(s(),y)) -> app(app(le(),x),y)
r6: app(app(maxlist(),x),app(app(cons(),y),ys)) -> app(app(if(),app(app(le(),x),y)),app(app(maxlist(),y),ys))
r7: app(app(maxlist(),x),nil()) -> x
r8: app(height(),app(app(node(),x),xs)) -> app(s(),app(app(maxlist(),|0|()),app(app(map(),height()),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 + x2 + (5,4)
            app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,3)
            map_A() = (2,2)
            cons_A() = (2,1)
            height_A() = (1,1)
            node_A() = (9,8)
    
      precedence:
      
        app# = app = map > height > cons > node
      
      partial status:
    
        pi(app#) = [1]
        pi(app) = []
        pi(map) = []
        pi(cons) = []
        pi(height) = []
        pi(node) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (2,1)
            app_A(x1,x2) = (1,0)
            map_A() = (5,4)
            cons_A() = (3,2)
            height_A() = (3,2)
            node_A() = (4,3)
    
      precedence:
      
        app = map > app# = cons = height = node
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(map) = []
        pi(cons) = []
        pi(height) = []
        pi(node) = []
    

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(map(),f),app(app(cons(),x),xs)) -> app#(f,x)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(le(),|0|()),y) -> true()
r4: app(app(le(),app(s(),x)),|0|()) -> false()
r5: app(app(le(),app(s(),x)),app(s(),y)) -> app(app(le(),x),y)
r6: app(app(maxlist(),x),app(app(cons(),y),ys)) -> app(app(if(),app(app(le(),x),y)),app(app(maxlist(),y),ys))
r7: app(app(maxlist(),x),nil()) -> x
r8: app(height(),app(app(node(),x),xs)) -> app(s(),app(app(maxlist(),|0|()),app(app(map(),height()),xs)))

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(map(),f),app(app(cons(),x),xs)) -> app#(f,x)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(le(),|0|()),y) -> true()
r4: app(app(le(),app(s(),x)),|0|()) -> false()
r5: app(app(le(),app(s(),x)),app(s(),y)) -> app(app(le(),x),y)
r6: app(app(maxlist(),x),app(app(cons(),y),ys)) -> app(app(if(),app(app(le(),x),y)),app(app(maxlist(),y),ys))
r7: app(app(maxlist(),x),nil()) -> x
r8: app(height(),app(app(node(),x),xs)) -> app(s(),app(app(maxlist(),|0|()),app(app(map(),height()),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) = ((1,0),(1,1)) x1 + (4,4)
            app_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (2,2)
            map_A() = (5,1)
            cons_A() = (1,1)
    
      precedence:
      
        app# = app = map = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(map) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (1,2)
            app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,3)
            map_A() = (2,1)
            cons_A() = (2,3)
    
      precedence:
      
        app# = app > map = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(map) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: app#(app(maxlist(),x),app(app(cons(),y),ys)) -> app#(app(maxlist(),y),ys)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(le(),|0|()),y) -> true()
r4: app(app(le(),app(s(),x)),|0|()) -> false()
r5: app(app(le(),app(s(),x)),app(s(),y)) -> app(app(le(),x),y)
r6: app(app(maxlist(),x),app(app(cons(),y),ys)) -> app(app(if(),app(app(le(),x),y)),app(app(maxlist(),y),ys))
r7: app(app(maxlist(),x),nil()) -> x
r8: app(height(),app(app(node(),x),xs)) -> app(s(),app(app(maxlist(),|0|()),app(app(map(),height()),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) = ((1,0),(0,0)) x2 + (1,2)
            app_A(x1,x2) = x2 + (4,2)
            maxlist_A() = (3,1)
            cons_A() = (2,3)
    
      precedence:
      
        app# = app = maxlist = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(maxlist) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (0,0)
            app_A(x1,x2) = x2 + (0,2)
            maxlist_A() = (1,3)
            cons_A() = (0,1)
    
      precedence:
      
        cons > app# = app = maxlist
      
      partial status:
    
        pi(app#) = []
        pi(app) = [2]
        pi(maxlist) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: app#(app(le(),app(s(),x)),app(s(),y)) -> app#(app(le(),x),y)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(app(le(),|0|()),y) -> true()
r4: app(app(le(),app(s(),x)),|0|()) -> false()
r5: app(app(le(),app(s(),x)),app(s(),y)) -> app(app(le(),x),y)
r6: app(app(maxlist(),x),app(app(cons(),y),ys)) -> app(app(if(),app(app(le(),x),y)),app(app(maxlist(),y),ys))
r7: app(app(maxlist(),x),nil()) -> x
r8: app(height(),app(app(node(),x),xs)) -> app(s(),app(app(maxlist(),|0|()),app(app(map(),height()),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) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (4,11)
            app_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (3,1)
            le_A() = (5,5)
            s_A() = (6,4)
    
      precedence:
      
        app# > le > app = s
      
      partial status:
    
        pi(app#) = [2]
        pi(app) = [2]
        pi(le) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((1,0),(1,1)) x2
            app_A(x1,x2) = x2 + (3,0)
            le_A() = (1,1)
            s_A() = (2,2)
    
      precedence:
      
        app# > app = le = s
      
      partial status:
    
        pi(app#) = []
        pi(app) = [2]
        pi(le) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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