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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1 + x2
        app_A(x1,x2) = x1 + x2
        map_A() = 1
        cons_A() = 1
        size_A() = 1
        node_A() = 2
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x2
        app_A(x1,x2) = 1
        map_A() = 1
        cons_A() = 0
        size_A() = 1
        node_A() = 0
    

The next rules are strictly ordered:

  p1, p2, p3

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1 + x2
        sum_A() = 0
        app_A(x1,x2) = x2 + 1
        cons_A() = 0
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1 + x2
        sum_A() = 0
        app_A(x1,x2) = x2 + 1
        cons_A() = 0
    

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1 + x2
        app_A(x1,x2) = x2 + 1
        plus_A() = 0
        s_A() = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1 + x2
        app_A(x1,x2) = x2 + 1
        plus_A() = 0
        s_A() = 1
    

The next rules are strictly ordered:

  p1

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