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(sum(),app(app(cons(),x),xs)) -> app(app(plus(),x),app(sum(),xs))
  app(size(),app(app(node(),x),xs)) -> app(s(),app(sum(),app(app(map(),size()),xs)))
  app(app(plus(),|0|()),x) -> |0|()
  app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))

-- 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#(sum(),app(app(cons(),x),xs)) -> app#(app(plus(),x),app(sum(),xs))
p6: app#(sum(),app(app(cons(),x),xs)) -> app#(plus(),x)
p7: app#(sum(),app(app(cons(),x),xs)) -> app#(sum(),xs)
p8: app#(size(),app(app(node(),x),xs)) -> app#(s(),app(sum(),app(app(map(),size()),xs)))
p9: app#(size(),app(app(node(),x),xs)) -> app#(sum(),app(app(map(),size()),xs))
p10: app#(size(),app(app(node(),x),xs)) -> app#(app(map(),size()),xs)
p11: app#(size(),app(app(node(),x),xs)) -> app#(map(),size())
p12: app#(app(plus(),app(s(),x)),y) -> app#(s(),app(app(plus(),x),y))
p13: app#(app(plus(),app(s(),x)),y) -> app#(app(plus(),x),y)
p14: app#(app(plus(),app(s(),x)),y) -> app#(plus(),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(sum(),app(app(cons(),x),xs)) -> app(app(plus(),x),app(sum(),xs))
r4: app(size(),app(app(node(),x),xs)) -> app(s(),app(sum(),app(app(map(),size()),xs)))
r5: app(app(plus(),|0|()),x) -> |0|()
r6: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p10}
  {p7}
  {p13}


-- 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#(size(),app(app(node(),x),xs)) -> app#(app(map(),size()),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(sum(),app(app(cons(),x),xs)) -> app(app(plus(),x),app(sum(),xs))
r4: app(size(),app(app(node(),x),xs)) -> app(s(),app(sum(),app(app(map(),size()),xs)))
r5: app(app(plus(),|0|()),x) -> |0|()
r6: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))

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{1, x1 - 12, x2 - 16}
        app_A(x1,x2) = max{26, x1 + 13, x2 + 18}
        map_A = 4
        cons_A = 5
        size_A = 3
        node_A = 13
  
    precedence:
    
      app# = app = map > cons = size = node
    
    partial status:
  
      pi(app#) = []
      pi(app) = [1, 2]
      pi(map) = []
      pi(cons) = []
      pi(size) = []
      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#(size(),app(app(node(),x),xs)) -> app#(app(map(),size()),xs)
p2: 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(sum(),app(app(cons(),x),xs)) -> app(app(plus(),x),app(sum(),xs))
r4: app(size(),app(app(node(),x),xs)) -> app(s(),app(sum(),app(app(map(),size()),xs)))
r5: app(app(plus(),|0|()),x) -> |0|()
r6: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))

The estimated dependency graph contains the following SCCs:

  {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#(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(sum(),app(app(cons(),x),xs)) -> app(app(plus(),x),app(sum(),xs))
r4: app(size(),app(app(node(),x),xs)) -> app(s(),app(sum(),app(app(map(),size()),xs)))
r5: app(app(plus(),|0|()),x) -> |0|()
r6: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))

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, x2 - 5}
        app_A(x1,x2) = max{x1 + 7, x2 + 4}
        map_A = 3
        cons_A = 4
  
    precedence:
    
      app# = app = map = cons
    
    partial status:
  
      pi(app#) = []
      pi(app) = [1, 2]
      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#(sum(),app(app(cons(),x),xs)) -> app#(sum(),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(sum(),app(app(cons(),x),xs)) -> app(app(plus(),x),app(sum(),xs))
r4: app(size(),app(app(node(),x),xs)) -> app(s(),app(sum(),app(app(map(),size()),xs)))
r5: app(app(plus(),|0|()),x) -> |0|()
r6: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))

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{x1 + 4, x2 + 7}
        sum_A = 1
        app_A(x1,x2) = max{x1 + 2, x2 + 2}
        cons_A = 0
  
    precedence:
    
      app# = sum = app = cons
    
    partial status:
  
      pi(app#) = []
      pi(sum) = []
      pi(app) = [2]
      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(plus(),app(s(),x)),y) -> app#(app(plus(),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(sum(),app(app(cons(),x),xs)) -> app(app(plus(),x),app(sum(),xs))
r4: app(size(),app(app(node(),x),xs)) -> app(s(),app(sum(),app(app(map(),size()),xs)))
r5: app(app(plus(),|0|()),x) -> |0|()
r6: app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))

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 - 9}
        app_A(x1,x2) = max{x1 + 2, x2 + 4}
        plus_A = 2
        s_A = 6
  
    precedence:
    
      app# = app = plus = s
    
    partial status:
  
      pi(app#) = []
      pi(app) = [1, 2]
      pi(plus) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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