YES
We show the termination of the TRS R:

  app(app(append(),nil()),ys) -> ys
  app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
  app(app(flatwith(),f),app(leaf(),x)) -> app(app(cons(),app(f,x)),nil())
  app(app(flatwith(),f),app(node(),xs)) -> app(app(flatwithsub(),f),xs)
  app(app(flatwithsub(),f),nil()) -> nil()
  app(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app(app(append(),app(app(flatwith(),f),x)),app(app(flatwithsub(),f),xs))

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(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(flatwith_0(),y) -> flatwith_1(y)
  app(flatwith_1(x1),y) -> flatwith_2(x1,y)
  app(flatwithsub_0(),y) -> flatwithsub_1(y)
  app(flatwithsub_1(x1),y) -> flatwithsub_2(x1,y)
  app(leaf_0(),y) -> leaf_1(y)
  app(node_0(),y) -> node_1(y)
  append_2(nil_0(),ys) -> ys
  append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
  flatwith_2(f,leaf_1(x)) -> cons_2(app(f,x),nil_0())
  flatwith_2(f,node_1(xs)) -> flatwithsub_2(f,xs)
  flatwithsub_2(f,nil_0()) -> nil_0()
  flatwithsub_2(f,cons_2(x,xs)) -> append_2(flatwith_2(f,x),flatwithsub_2(f,xs))


-- 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#(flatwith_1(x1),y) -> flatwith_2#(x1,y)
p3: app#(flatwithsub_1(x1),y) -> flatwithsub_2#(x1,y)
p4: append_2#(cons_2(x,xs),ys) -> append_2#(xs,ys)
p5: flatwith_2#(f,leaf_1(x)) -> app#(f,x)
p6: flatwith_2#(f,node_1(xs)) -> flatwithsub_2#(f,xs)
p7: flatwithsub_2#(f,cons_2(x,xs)) -> append_2#(flatwith_2(f,x),flatwithsub_2(f,xs))
p8: flatwithsub_2#(f,cons_2(x,xs)) -> flatwith_2#(f,x)
p9: flatwithsub_2#(f,cons_2(x,xs)) -> flatwithsub_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(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(flatwith_0(),y) -> flatwith_1(y)
r6: app(flatwith_1(x1),y) -> flatwith_2(x1,y)
r7: app(flatwithsub_0(),y) -> flatwithsub_1(y)
r8: app(flatwithsub_1(x1),y) -> flatwithsub_2(x1,y)
r9: app(leaf_0(),y) -> leaf_1(y)
r10: app(node_0(),y) -> node_1(y)
r11: append_2(nil_0(),ys) -> ys
r12: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r13: flatwith_2(f,leaf_1(x)) -> cons_2(app(f,x),nil_0())
r14: flatwith_2(f,node_1(xs)) -> flatwithsub_2(f,xs)
r15: flatwithsub_2(f,nil_0()) -> nil_0()
r16: flatwithsub_2(f,cons_2(x,xs)) -> append_2(flatwith_2(f,x),flatwithsub_2(f,xs))

The estimated dependency graph contains the following SCCs:

  {p2, p3, p5, p6, p8, p9}
  {p4}


-- Reduction pair.

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

p1: flatwith_2#(f,leaf_1(x)) -> app#(f,x)
p2: app#(flatwithsub_1(x1),y) -> flatwithsub_2#(x1,y)
p3: flatwithsub_2#(f,cons_2(x,xs)) -> flatwithsub_2#(f,xs)
p4: flatwithsub_2#(f,cons_2(x,xs)) -> flatwith_2#(f,x)
p5: flatwith_2#(f,node_1(xs)) -> flatwithsub_2#(f,xs)
p6: app#(flatwith_1(x1),y) -> flatwith_2#(x1,y)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(flatwith_0(),y) -> flatwith_1(y)
r6: app(flatwith_1(x1),y) -> flatwith_2(x1,y)
r7: app(flatwithsub_0(),y) -> flatwithsub_1(y)
r8: app(flatwithsub_1(x1),y) -> flatwithsub_2(x1,y)
r9: app(leaf_0(),y) -> leaf_1(y)
r10: app(node_0(),y) -> node_1(y)
r11: append_2(nil_0(),ys) -> ys
r12: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r13: flatwith_2(f,leaf_1(x)) -> cons_2(app(f,x),nil_0())
r14: flatwith_2(f,node_1(xs)) -> flatwithsub_2(f,xs)
r15: flatwithsub_2(f,nil_0()) -> nil_0()
r16: flatwithsub_2(f,cons_2(x,xs)) -> append_2(flatwith_2(f,x),flatwithsub_2(f,xs))

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              flatwith_2#_A(x1,x2) = x1 + x2
              leaf_1_A(x1) = x1 + 3
              app#_A(x1,x2) = x1 + x2
              flatwithsub_1_A(x1) = x1 + 5
              flatwithsub_2#_A(x1,x2) = x1 + x2
              cons_2_A(x1,x2) = x1 + x2 + 5
              node_1_A(x1) = x1 + 3
              flatwith_1_A(x1) = x1 + 6
              flatwith_2#!_A(x1,x2) = x1 + x2 + 6
              leaf_1!_A(x1) = 2
              app#!_A(x1,x2) = x1 + x2 + 1
              flatwithsub_1!_A(x1) = 5
              flatwithsub_2#!_A(x1,x2) = x1 + x2 + 6
              cons_2!_A(x1,x2) = 5
              node_1!_A(x1) = 1
              flatwith_1!_A(x1) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            flatwith_2#_A(x1,x2) = max{1, x2 - 48}
            leaf_1_A(x1) = x1 + 13
            app#_A(x1,x2) = max{1, x2 - 35}
            flatwithsub_1_A(x1) = x1 + 8
            flatwithsub_2#_A(x1,x2) = max{1, x2 - 35}
            cons_2_A(x1,x2) = max{x1 - 13, x2 + 34}
            node_1_A(x1) = x1 + 22
            flatwith_1_A(x1) = 8
            flatwith_2#!_A(x1,x2) = max{13, x1 + 2, x2 - 9}
            leaf_1!_A(x1) = 17
            app#!_A(x1,x2) = max{x1 - 9, x2 + 12}
            flatwithsub_1!_A(x1) = max{2, x1 + 1}
            flatwithsub_2#!_A(x1,x2) = 12
            cons_2!_A(x1,x2) = max{19, x1 + 11, x2 - 74}
            node_1!_A(x1) = max{0, x1 - 73}
            flatwith_1!_A(x1) = max{14, x1 - 1}
        
  
    partial status:
  
      pi(flatwith_2#) = []
      pi(leaf_1) = []
      pi(app#) = []
      pi(flatwithsub_1) = []
      pi(flatwithsub_2#) = []
      pi(cons_2) = [1, 2]
      pi(node_1) = []
      pi(flatwith_1) = [1]

The following rules are removed:

  p1, p3, p4, p5, p6



-- SCC decomposition.

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

p1: app#(flatwithsub_1(x1),y) -> flatwithsub_2#(x1,y)

and R consists of:

r1: app(append_0(),y) -> append_1(y)
r2: app(append_1(x1),y) -> append_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(flatwith_0(),y) -> flatwith_1(y)
r6: app(flatwith_1(x1),y) -> flatwith_2(x1,y)
r7: app(flatwithsub_0(),y) -> flatwithsub_1(y)
r8: app(flatwithsub_1(x1),y) -> flatwithsub_2(x1,y)
r9: app(leaf_0(),y) -> leaf_1(y)
r10: app(node_0(),y) -> node_1(y)
r11: append_2(nil_0(),ys) -> ys
r12: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r13: flatwith_2(f,leaf_1(x)) -> cons_2(app(f,x),nil_0())
r14: flatwith_2(f,node_1(xs)) -> flatwithsub_2(f,xs)
r15: flatwithsub_2(f,nil_0()) -> nil_0()
r16: flatwithsub_2(f,cons_2(x,xs)) -> append_2(flatwith_2(f,x),flatwithsub_2(f,xs))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(flatwith_0(),y) -> flatwith_1(y)
r6: app(flatwith_1(x1),y) -> flatwith_2(x1,y)
r7: app(flatwithsub_0(),y) -> flatwithsub_1(y)
r8: app(flatwithsub_1(x1),y) -> flatwithsub_2(x1,y)
r9: app(leaf_0(),y) -> leaf_1(y)
r10: app(node_0(),y) -> node_1(y)
r11: append_2(nil_0(),ys) -> ys
r12: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r13: flatwith_2(f,leaf_1(x)) -> cons_2(app(f,x),nil_0())
r14: flatwith_2(f,node_1(xs)) -> flatwithsub_2(f,xs)
r15: flatwithsub_2(f,nil_0()) -> nil_0()
r16: flatwithsub_2(f,cons_2(x,xs)) -> append_2(flatwith_2(f,x),flatwithsub_2(f,xs))

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!):
        
          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) = x2 + 2
              append_2#!_A(x1,x2) = x1 + 3
              cons_2!_A(x1,x2) = x1 + 2
        
        2. 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{1, x2 - 11}
            append_2#!_A(x1,x2) = max{4, x2 + 1}
            cons_2!_A(x1,x2) = max{3, x2 - 6}
        
  
    partial status:
  
      pi(append_2#) = []
      pi(cons_2) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.