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(flatten(),app(app(node(),x),xs)) -> app(app(cons(),x),app(concat(),app(app(map(),flatten()),xs)))
  app(concat(),nil()) -> nil()
  app(concat(),app(app(cons(),x),xs)) -> app(app(append(),x),app(concat(),xs))
  app(app(append(),nil()),xs) -> xs
  app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))

Instead we prove termination of the uncurried system:

  app(append_0(),y) -> append_1(y)
  app(append_1(x1),y) -> append_2(x1,y)
  app(concat_0(),y) -> concat_1(y)
  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(flatten_0(),y) -> flatten_1(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)
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  flatten_1(node_2(x,xs)) -> cons_2(x,concat_1(map_2(flatten_0(),xs)))
  concat_1(nil_0()) -> nil_0()
  concat_1(cons_2(x,xs)) -> append_2(x,concat_1(xs))
  append_2(nil_0(),xs) -> xs
  append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))


-- SCC decomposition.

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

p1: app#(append_1(x1),y) -> append_2#(x1,y)
p2: app#(concat_0(),y) -> concat_1#(y)
p3: app#(flatten_0(),y) -> flatten_1#(y)
p4: app#(map_1(x1),y) -> map_2#(x1,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: flatten_1#(node_2(x,xs)) -> concat_1#(map_2(flatten_0(),xs))
p8: flatten_1#(node_2(x,xs)) -> map_2#(flatten_0(),xs)
p9: concat_1#(cons_2(x,xs)) -> append_2#(x,concat_1(xs))
p10: concat_1#(cons_2(x,xs)) -> concat_1#(xs)
p11: append_2#(cons_2(x,xs),ys) -> append_2#(xs,ys)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(concat_0(),y) -> concat_1(y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(flatten_0(),y) -> flatten_1(y)
r7: app(map_0(),y) -> map_1(y)
r8: app(map_1(x1),y) -> map_2(x1,y)
r9: app(node_0(),y) -> node_1(y)
r10: app(node_1(x1),y) -> node_2(x1,y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: flatten_1(node_2(x,xs)) -> cons_2(x,concat_1(map_2(flatten_0(),xs)))
r14: concat_1(nil_0()) -> nil_0()
r15: concat_1(cons_2(x,xs)) -> append_2(x,concat_1(xs))
r16: append_2(nil_0(),xs) -> xs
r17: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p5, p6, p8}
  {p10}
  {p11}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p4: app#(flatten_0(),y) -> flatten_1#(y)
p5: flatten_1#(node_2(x,xs)) -> map_2#(flatten_0(),xs)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(concat_0(),y) -> concat_1(y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(flatten_0(),y) -> flatten_1(y)
r7: app(map_0(),y) -> map_1(y)
r8: app(map_1(x1),y) -> map_2(x1,y)
r9: app(node_0(),y) -> node_1(y)
r10: app(node_1(x1),y) -> node_2(x1,y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: flatten_1(node_2(x,xs)) -> cons_2(x,concat_1(map_2(flatten_0(),xs)))
r14: concat_1(nil_0()) -> nil_0()
r15: concat_1(cons_2(x,xs)) -> append_2(x,concat_1(xs))
r16: append_2(nil_0(),xs) -> xs
r17: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))

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) = max{1, x2 - 15}
            cons_2_A(x1,x2) = max{x1 + 14, x2}
            app#_A(x1,x2) = max{1, x2 - 1}
            map_1_A(x1) = max{2, x1 - 2}
            flatten_0_A = 0
            flatten_1#_A(x1) = max{1, x1 - 42}
            node_2_A(x1,x2) = max{x1 + 40, x2 + 41}
            map_2#!_A(x1,x2) = 1
            cons_2!_A(x1,x2) = max{x1 + 27, x2 + 8}
            app#!_A(x1,x2) = 1
            map_1!_A(x1) = 2
            flatten_0!_A = 2
            flatten_1#!_A(x1) = 1
            node_2!_A(x1,x2) = max{3, x1 - 3, x2 - 40}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x1 + x2 + 5
              app#_A(x1,x2) = x2
              map_1_A(x1) = x1 + 3
              flatten_0_A() = 1
              flatten_1#_A(x1) = x1
              node_2_A(x1,x2) = x1 + x2 + 5
              map_2#!_A(x1,x2) = x1 + 1
              cons_2!_A(x1,x2) = x1 + 5
              app#!_A(x1,x2) = x1 + 1
              map_1!_A(x1) = x1 + 2
              flatten_0!_A() = 3
              flatten_1#!_A(x1) = 2
              node_2!_A(x1,x2) = 4
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = [1]
      pi(app#) = []
      pi(map_1) = [1]
      pi(flatten_0) = []
      pi(flatten_1#) = []
      pi(node_2) = [1, 2]

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p3: app#(flatten_0(),y) -> flatten_1#(y)
p4: flatten_1#(node_2(x,xs)) -> map_2#(flatten_0(),xs)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(concat_0(),y) -> concat_1(y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(flatten_0(),y) -> flatten_1(y)
r7: app(map_0(),y) -> map_1(y)
r8: app(map_1(x1),y) -> map_2(x1,y)
r9: app(node_0(),y) -> node_1(y)
r10: app(node_1(x1),y) -> node_2(x1,y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: flatten_1(node_2(x,xs)) -> cons_2(x,concat_1(map_2(flatten_0(),xs)))
r14: concat_1(nil_0()) -> nil_0()
r15: concat_1(cons_2(x,xs)) -> append_2(x,concat_1(xs))
r16: append_2(nil_0(),xs) -> xs
r17: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(flatten_0(),y) -> flatten_1#(y)
p3: flatten_1#(node_2(x,xs)) -> map_2#(flatten_0(),xs)
p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(concat_0(),y) -> concat_1(y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(flatten_0(),y) -> flatten_1(y)
r7: app(map_0(),y) -> map_1(y)
r8: app(map_1(x1),y) -> map_2(x1,y)
r9: app(node_0(),y) -> node_1(y)
r10: app(node_1(x1),y) -> node_2(x1,y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: flatten_1(node_2(x,xs)) -> cons_2(x,concat_1(map_2(flatten_0(),xs)))
r14: concat_1(nil_0()) -> nil_0()
r15: concat_1(cons_2(x,xs)) -> append_2(x,concat_1(xs))
r16: append_2(nil_0(),xs) -> xs
r17: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))

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) = max{0, x1 - 1, x2 - 13}
            cons_2_A(x1,x2) = max{4, x1 - 2, x2 + 1}
            app#_A(x1,x2) = max{0, x1 - 1, x2 - 31}
            flatten_0_A = 0
            flatten_1#_A(x1) = max{0, x1 - 31}
            node_2_A(x1,x2) = max{29, x1 + 4, x2 + 18}
            map_2#!_A(x1,x2) = max{x1 + 9, x2 - 5}
            cons_2!_A(x1,x2) = max{9, x2 - 11}
            app#!_A(x1,x2) = max{8, x1 - 1, x2 - 8}
            flatten_0!_A = 4
            flatten_1#!_A(x1) = max{7, x1 - 19}
            node_2!_A(x1,x2) = max{5, x1 - 1, x2 + 3}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x1 + x2 + 5
              app#_A(x1,x2) = x2
              flatten_0_A() = 1
              flatten_1#_A(x1) = x1
              node_2_A(x1,x2) = x1 + x2 + 5
              map_2#!_A(x1,x2) = x2
              cons_2!_A(x1,x2) = x1 + x2 + 5
              app#!_A(x1,x2) = x2 + 2
              flatten_0!_A() = 0
              flatten_1#!_A(x1) = x1 + 1
              node_2!_A(x1,x2) = 1
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = [1]
      pi(app#) = []
      pi(flatten_0) = []
      pi(flatten_1#) = []
      pi(node_2) = [1]

The following rules are removed:

  p1, p2, p3, p4

Then no dependency pair remains.

-- Reduction pair.

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

p1: concat_1#(cons_2(x,xs)) -> concat_1#(xs)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(concat_0(),y) -> concat_1(y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(flatten_0(),y) -> flatten_1(y)
r7: app(map_0(),y) -> map_1(y)
r8: app(map_1(x1),y) -> map_2(x1,y)
r9: app(node_0(),y) -> node_1(y)
r10: app(node_1(x1),y) -> node_2(x1,y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: flatten_1(node_2(x,xs)) -> cons_2(x,concat_1(map_2(flatten_0(),xs)))
r14: concat_1(nil_0()) -> nil_0()
r15: concat_1(cons_2(x,xs)) -> append_2(x,concat_1(xs))
r16: append_2(nil_0(),xs) -> xs
r17: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))

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:
          
            concat_1#_A(x1) = 0
            cons_2_A(x1,x2) = max{2, x1 - 26, x2 - 27}
            concat_1#!_A(x1) = 10
            cons_2!_A(x1,x2) = x1 + 16
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              concat_1#_A(x1) = x1
              cons_2_A(x1,x2) = x1 + x2 + 5
              concat_1#!_A(x1) = x1 + 11
              cons_2!_A(x1,x2) = 10
        
  
    partial status:
  
      pi(concat_1#) = []
      pi(cons_2) = [1]

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: append_2#(cons_2(x,xs),ys) -> append_2#(xs,ys)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(concat_0(),y) -> concat_1(y)
r4: app(cons_0(),y) -> cons_1(y)
r5: app(cons_1(x1),y) -> cons_2(x1,y)
r6: app(flatten_0(),y) -> flatten_1(y)
r7: app(map_0(),y) -> map_1(y)
r8: app(map_1(x1),y) -> map_2(x1,y)
r9: app(node_0(),y) -> node_1(y)
r10: app(node_1(x1),y) -> node_2(x1,y)
r11: map_2(f,nil_0()) -> nil_0()
r12: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r13: flatten_1(node_2(x,xs)) -> cons_2(x,concat_1(map_2(flatten_0(),xs)))
r14: concat_1(nil_0()) -> nil_0()
r15: concat_1(cons_2(x,xs)) -> append_2(x,concat_1(xs))
r16: append_2(nil_0(),xs) -> xs
r17: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))

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:
          
            append_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{3, x2 - 8}
            append_2#!_A(x1,x2) = max{4, x2 + 1}
            cons_2!_A(x1,x2) = max{3, x2 - 8}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              append_2#_A(x1,x2) = x1
              cons_2_A(x1,x2) = x1 + x2 + 5
              append_2#!_A(x1,x2) = x1 + 7
              cons_2!_A(x1,x2) = x1 + x2 + 6
        
  
    partial status:
  
      pi(append_2#) = []
      pi(cons_2) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.