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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (12,3)
          app_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (4,4)
          map_A() = (2,2)
          cons_A() = (3,8)
          height_A() = (3,3)
          node_A() = (0,1)
  
    precedence:
    
      app# = app = map = cons = 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:

  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),(0,0)) x2 + (10,5)
          app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (4,3)
          map_A() = (1,4)
          cons_A() = (1,6)
          height_A() = (2,2)
          node_A() = (0,1)
  
    precedence:
    
      app# = app = map = 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = ((1,0),(0,0)) x2
          app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,1)
          map_A() = (1,1)
          cons_A() = (1,1)
  
    precedence:
    
      app = map > app# = 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = ((0,0),(1,0)) x2 + (5,4)
          app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,3)
          maxlist_A() = (1,1)
          cons_A() = (3,2)
  
    precedence:
    
      app# = app = maxlist = cons
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = x1 + (3,1)
          app_A(x1,x2) = ((1,0),(1,1)) x2 + (3,0)
          le_A() = (1,2)
          s_A() = (2,3)
  
    precedence:
    
      app# = app = le = s
    
    partial status:
  
      pi(app#) = [1]
      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.