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))

Instead we prove termination of the uncurried system:

  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(node_0(),y) -> node_1(y)
  app(node_1(x1),y) -> node_2(x1,y)
  app(plus_0(),y) -> plus_1(y)
  app(plus_1(x1),y) -> plus_2(x1,y)
  app(s_0(),y) -> s_1(y)
  app(size_0(),y) -> size_1(y)
  app(sum_0(),y) -> sum_1(y)
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  sum_1(cons_2(x,xs)) -> plus_2(x,sum_1(xs))
  size_1(node_2(x,xs)) -> s_1(sum_1(map_2(size_0(),xs)))
  plus_2(|0|_0(),x) -> |0|_0()
  plus_2(s_1(x),y) -> s_1(plus_2(x,y))


-- SCC decomposition.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)
p2: app#(plus_1(x1),y) -> plus_2#(x1,y)
p3: app#(size_0(),y) -> size_1#(y)
p4: app#(sum_0(),y) -> sum_1#(y)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p7: sum_1#(cons_2(x,xs)) -> plus_2#(x,sum_1(xs))
p8: sum_1#(cons_2(x,xs)) -> sum_1#(xs)
p9: size_1#(node_2(x,xs)) -> sum_1#(map_2(size_0(),xs))
p10: size_1#(node_2(x,xs)) -> map_2#(size_0(),xs)
p11: plus_2#(s_1(x),y) -> plus_2#(x,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(node_0(),y) -> node_1(y)
r6: app(node_1(x1),y) -> node_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(size_0(),y) -> size_1(y)
r11: app(sum_0(),y) -> sum_1(y)
r12: map_2(f,nil_0()) -> nil_0()
r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r14: sum_1(cons_2(x,xs)) -> plus_2(x,sum_1(xs))
r15: size_1(node_2(x,xs)) -> s_1(sum_1(map_2(size_0(),xs)))
r16: plus_2(|0|_0(),x) -> |0|_0()
r17: plus_2(s_1(x),y) -> s_1(plus_2(x,y))

The estimated dependency graph contains the following SCCs:

  {p1, p3, p5, p6, p10}
  {p8}
  {p11}


-- Reduction pair.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)
p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p3: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p4: app#(size_0(),y) -> size_1#(y)
p5: size_1#(node_2(x,xs)) -> map_2#(size_0(),xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(node_0(),y) -> node_1(y)
r6: app(node_1(x1),y) -> node_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(size_0(),y) -> size_1(y)
r11: app(sum_0(),y) -> sum_1(y)
r12: map_2(f,nil_0()) -> nil_0()
r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r14: sum_1(cons_2(x,xs)) -> plus_2(x,sum_1(xs))
r15: size_1(node_2(x,xs)) -> s_1(sum_1(map_2(size_0(),xs)))
r16: plus_2(|0|_0(),x) -> |0|_0()
r17: plus_2(s_1(x),y) -> s_1(plus_2(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 14, x2 - 37}
            map_1_A(x1) = max{13, x1 + 6}
            map_2#_A(x1,x2) = max{0, x1 - 8, x2 - 45}
            cons_2_A(x1,x2) = max{10, x1 + 8, x2 + 1}
            size_0_A = 1
            size_1#_A(x1) = max{0, x1 - 39}
            node_2_A(x1,x2) = max{31, x1 - 10, x2 - 6}
            app#!_A(x1,x2) = max{4, x1 - 7, x2 - 49}
            map_1!_A(x1) = max{11, x1 - 33}
            map_2#!_A(x1,x2) = max{5, x1 - 2, x2 - 56}
            cons_2!_A(x1,x2) = max{3, x1 - 88, x2 - 56}
            size_0!_A = 2
            size_1#!_A(x1) = max{3, x1 - 50}
            node_2!_A(x1,x2) = max{6, x1 - 2, x2 - 38}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 47, x2 - 5}
            map_1_A(x1) = max{46, x1 + 37}
            map_2#_A(x1,x2) = max{0, x1 - 38, x2 - 7}
            cons_2_A(x1,x2) = max{6, x1 + 2, x2}
            size_0_A = 37
            size_1#_A(x1) = max{0, x1 - 5}
            node_2_A(x1,x2) = max{x1, x2}
            app#!_A(x1,x2) = max{3, x1 - 45, x2 - 52}
            map_1!_A(x1) = 2
            map_2#!_A(x1,x2) = max{5, x1 - 38, x2 - 51}
            cons_2!_A(x1,x2) = max{2, x1 - 38, x2 - 36}
            size_0!_A = 1
            size_1#!_A(x1) = max{2, x1 - 53}
            node_2!_A(x1,x2) = max{6, x1 - 2, x2 - 50}
        
  
    partial status:
  
      pi(app#) = [2]
      pi(map_1) = [1]
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(size_0) = []
      pi(size_1#) = [1]
      pi(node_2) = [2]

The following rules are removed:

  p1, p3, p4, p5



-- SCC decomposition.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(node_0(),y) -> node_1(y)
r6: app(node_1(x1),y) -> node_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(size_0(),y) -> size_1(y)
r11: app(sum_0(),y) -> sum_1(y)
r12: map_2(f,nil_0()) -> nil_0()
r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r14: sum_1(cons_2(x,xs)) -> plus_2(x,sum_1(xs))
r15: size_1(node_2(x,xs)) -> s_1(sum_1(map_2(size_0(),xs)))
r16: plus_2(|0|_0(),x) -> |0|_0()
r17: plus_2(s_1(x),y) -> s_1(plus_2(x,y))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(node_0(),y) -> node_1(y)
r6: app(node_1(x1),y) -> node_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(size_0(),y) -> size_1(y)
r11: app(sum_0(),y) -> sum_1(y)
r12: map_2(f,nil_0()) -> nil_0()
r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r14: sum_1(cons_2(x,xs)) -> plus_2(x,sum_1(xs))
r15: size_1(node_2(x,xs)) -> s_1(sum_1(map_2(size_0(),xs)))
r16: plus_2(|0|_0(),x) -> |0|_0()
r17: plus_2(s_1(x),y) -> s_1(plus_2(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{x1 + 3, x2 - 16}
            map_2#!_A(x1,x2) = max{20, x2 + 17}
            cons_2!_A(x1,x2) = max{x1 - 28, x2 + 21}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{3, x1 - 2, x2 - 12}
            map_2#!_A(x1,x2) = max{9, x1 - 1, x2 - 4}
            cons_2!_A(x1,x2) = max{8, x1 - 52, x2 - 5}
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: sum_1#(cons_2(x,xs)) -> sum_1#(xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(node_0(),y) -> node_1(y)
r6: app(node_1(x1),y) -> node_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(size_0(),y) -> size_1(y)
r11: app(sum_0(),y) -> sum_1(y)
r12: map_2(f,nil_0()) -> nil_0()
r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r14: sum_1(cons_2(x,xs)) -> plus_2(x,sum_1(xs))
r15: size_1(node_2(x,xs)) -> s_1(sum_1(map_2(size_0(),xs)))
r16: plus_2(|0|_0(),x) -> |0|_0()
r17: plus_2(s_1(x),y) -> s_1(plus_2(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum_1#_A(x1) = 0
            cons_2_A(x1,x2) = max{3, x1 - 2, x2 - 4}
            sum_1#!_A(x1) = max{1, x1 - 39}
            cons_2!_A(x1,x2) = max{3, x1 - 44, x2 - 39}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum_1#_A(x1) = 0
            cons_2_A(x1,x2) = max{3, x1 - 2, x2 - 13}
            sum_1#!_A(x1) = max{1, x1 - 93}
            cons_2!_A(x1,x2) = max{9, x1 - 95, x2 - 93}
        
  
    partial status:
  
      pi(sum_1#) = [1]
      pi(cons_2) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: plus_2#(s_1(x),y) -> plus_2#(x,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(map_0(),y) -> map_1(y)
r4: app(map_1(x1),y) -> map_2(x1,y)
r5: app(node_0(),y) -> node_1(y)
r6: app(node_1(x1),y) -> node_2(x1,y)
r7: app(plus_0(),y) -> plus_1(y)
r8: app(plus_1(x1),y) -> plus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: app(size_0(),y) -> size_1(y)
r11: app(sum_0(),y) -> sum_1(y)
r12: map_2(f,nil_0()) -> nil_0()
r13: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r14: sum_1(cons_2(x,xs)) -> plus_2(x,sum_1(xs))
r15: size_1(node_2(x,xs)) -> s_1(sum_1(map_2(size_0(),xs)))
r16: plus_2(|0|_0(),x) -> |0|_0()
r17: plus_2(s_1(x),y) -> s_1(plus_2(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{4, x1 - 4}
            plus_2#!_A(x1,x2) = max{0, x1 - 7}
            s_1!_A(x1) = max{6, x1 - 6}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{4, x1 - 4}
            plus_2#!_A(x1,x2) = max{1, x1 - 5, x2 - 1}
            s_1!_A(x1) = max{7, x1 - 19}
        
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.