YES
We show the termination of the TRS R:

  app(app(neq(),|0|()),|0|()) -> false()
  app(app(neq(),|0|()),app(s(),y)) -> true()
  app(app(neq(),app(s(),x)),|0|()) -> true()
  app(app(neq(),app(s(),x)),app(s(),y)) -> app(app(neq(),x),y)
  app(app(filter(),f),nil()) -> nil()
  app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
  app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys))
  app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys)
  nonzero() -> app(filter(),app(neq(),|0|()))

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(filter_0(),y) -> filter_1(y)
  app(filter_1(x1),y) -> filter_2(x1,y)
  app(filtersub_0(),y) -> filtersub_1(y)
  app(filtersub_1(x1),y) -> filtersub_2(x1,y)
  app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y)
  app(neq_0(),y) -> neq_1(y)
  app(neq_1(x1),y) -> neq_2(x1,y)
  app(s_0(),y) -> s_1(y)
  neq_2(|0|_0(),|0|_0()) -> false_0()
  neq_2(|0|_0(),s_1(y)) -> true_0()
  neq_2(s_1(x),|0|_0()) -> true_0()
  neq_2(s_1(x),s_1(y)) -> neq_2(x,y)
  filter_2(f,nil_0()) -> nil_0()
  filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys))
  filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys))
  filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys)
  nonzero_0() -> filter_1(neq_1(|0|_0()))


-- SCC decomposition.

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

p1: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: app#(filtersub_2(x1,x2),y) -> filtersub_3#(x1,x2,y)
p3: app#(neq_1(x1),y) -> neq_2#(x1,y)
p4: neq_2#(s_1(x),s_1(y)) -> neq_2#(x,y)
p5: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys))
p6: filter_2#(f,cons_2(y,ys)) -> app#(f,y)
p7: filtersub_3#(true_0(),f,cons_2(y,ys)) -> filter_2#(f,ys)
p8: filtersub_3#(false_0(),f,cons_2(y,ys)) -> filter_2#(f,ys)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filtersub_0(),y) -> filtersub_1(y)
r6: app(filtersub_1(x1),y) -> filtersub_2(x1,y)
r7: app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y)
r8: app(neq_0(),y) -> neq_1(y)
r9: app(neq_1(x1),y) -> neq_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: neq_2(|0|_0(),|0|_0()) -> false_0()
r12: neq_2(|0|_0(),s_1(y)) -> true_0()
r13: neq_2(s_1(x),|0|_0()) -> true_0()
r14: neq_2(s_1(x),s_1(y)) -> neq_2(x,y)
r15: filter_2(f,nil_0()) -> nil_0()
r16: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys))
r17: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys))
r18: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys)
r19: nonzero_0() -> filter_1(neq_1(|0|_0()))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p5, p6, p7, p8}
  {p4}


-- Reduction pair.

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

p1: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: filter_2#(f,cons_2(y,ys)) -> app#(f,y)
p3: app#(filtersub_2(x1,x2),y) -> filtersub_3#(x1,x2,y)
p4: filtersub_3#(false_0(),f,cons_2(y,ys)) -> filter_2#(f,ys)
p5: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys))
p6: filtersub_3#(true_0(),f,cons_2(y,ys)) -> filter_2#(f,ys)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filtersub_0(),y) -> filtersub_1(y)
r6: app(filtersub_1(x1),y) -> filtersub_2(x1,y)
r7: app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y)
r8: app(neq_0(),y) -> neq_1(y)
r9: app(neq_1(x1),y) -> neq_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: neq_2(|0|_0(),|0|_0()) -> false_0()
r12: neq_2(|0|_0(),s_1(y)) -> true_0()
r13: neq_2(s_1(x),|0|_0()) -> true_0()
r14: neq_2(s_1(x),s_1(y)) -> neq_2(x,y)
r15: filter_2(f,nil_0()) -> nil_0()
r16: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys))
r17: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys))
r18: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys)
r19: nonzero_0() -> filter_1(neq_1(|0|_0()))

The set of usable rules consists of

  r11, r12, r13, r14, r15

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{1, x1 - 12, x2 - 3}
            filter_1_A(x1) = max{11, x1}
            filter_2#_A(x1,x2) = max{1, x1 - 12, x2 - 11}
            cons_2_A(x1,x2) = max{10, x1 + 8, x2 + 8}
            filtersub_2_A(x1,x2) = max{10, x1 + 8, x2 + 7}
            filtersub_3#_A(x1,x2,x3) = max{1, x2 - 12, x3 - 11}
            false_0_A = 20
            app_A(x1,x2) = max{20, x1 - 14}
            true_0_A = 20
            neq_2_A(x1,x2) = max{20, x1 - 15, x2 - 2}
            |0|_0_A = 0
            s_1_A(x1) = max{1, x1}
            filter_2_A(x1,x2) = max{12, x1 - 15, x2 + 1}
            nil_0_A = 13
            app#!_A(x1,x2) = max{0, x2 - 1}
            filter_1!_A(x1) = max{10, x1 - 14}
            filter_2#!_A(x1,x2) = max{0, x2 - 4}
            cons_2!_A(x1,x2) = 3
            filtersub_2!_A(x1,x2) = max{9, x1 + 3}
            filtersub_3#!_A(x1,x2,x3) = max{0, x3 - 11}
            false_0!_A = 10
            app!_A(x1,x2) = max{x1 - 1, x2 + 4}
            true_0!_A = 9
            neq_2!_A(x1,x2) = max{11, x1 - 4, x2 - 2}
            |0|_0!_A = 12
            s_1!_A(x1) = max{10, x1 - 3}
            filter_2!_A(x1,x2) = 9
            nil_0!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = x1 + x2
              filter_1_A(x1) = x1 + 4
              filter_2#_A(x1,x2) = x1 + x2
              cons_2_A(x1,x2) = x1 + x2 + 3
              filtersub_2_A(x1,x2) = x1 + x2 + 1
              filtersub_3#_A(x1,x2,x3) = x2 + x3
              false_0_A() = 1
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              neq_2_A(x1,x2) = x2
              |0|_0_A() = 1
              s_1_A(x1) = x1 + 1
              filter_2_A(x1,x2) = x1 + x2 + 1
              nil_0_A() = 1
              app#!_A(x1,x2) = x1 + x2
              filter_1!_A(x1) = 8
              filter_2#!_A(x1,x2) = x2 + 3
              cons_2!_A(x1,x2) = 5
              filtersub_2!_A(x1,x2) = 7
              filtersub_3#!_A(x1,x2,x3) = x3 + 1
              false_0!_A() = 8
              app!_A(x1,x2) = x2 + 9
              true_0!_A() = 7
              neq_2!_A(x1,x2) = x1 + x2 + 11
              |0|_0!_A() = 12
              s_1!_A(x1) = 10
              filter_2!_A(x1,x2) = 7
              nil_0!_A() = 0
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(filtersub_2) = [1]
      pi(filtersub_3#) = []
      pi(false_0) = []
      pi(app) = [1]
      pi(true_0) = []
      pi(neq_2) = []
      pi(|0|_0) = []
      pi(s_1) = [1]
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p4, p5, p6



-- SCC decomposition.

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

p1: app#(filtersub_2(x1,x2),y) -> filtersub_3#(x1,x2,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(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filtersub_0(),y) -> filtersub_1(y)
r6: app(filtersub_1(x1),y) -> filtersub_2(x1,y)
r7: app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y)
r8: app(neq_0(),y) -> neq_1(y)
r9: app(neq_1(x1),y) -> neq_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: neq_2(|0|_0(),|0|_0()) -> false_0()
r12: neq_2(|0|_0(),s_1(y)) -> true_0()
r13: neq_2(s_1(x),|0|_0()) -> true_0()
r14: neq_2(s_1(x),s_1(y)) -> neq_2(x,y)
r15: filter_2(f,nil_0()) -> nil_0()
r16: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys))
r17: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys))
r18: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys)
r19: nonzero_0() -> filter_1(neq_1(|0|_0()))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: neq_2#(s_1(x),s_1(y)) -> neq_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(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filtersub_0(),y) -> filtersub_1(y)
r6: app(filtersub_1(x1),y) -> filtersub_2(x1,y)
r7: app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y)
r8: app(neq_0(),y) -> neq_1(y)
r9: app(neq_1(x1),y) -> neq_2(x1,y)
r10: app(s_0(),y) -> s_1(y)
r11: neq_2(|0|_0(),|0|_0()) -> false_0()
r12: neq_2(|0|_0(),s_1(y)) -> true_0()
r13: neq_2(s_1(x),|0|_0()) -> true_0()
r14: neq_2(s_1(x),s_1(y)) -> neq_2(x,y)
r15: filter_2(f,nil_0()) -> nil_0()
r16: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys))
r17: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys))
r18: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys)
r19: nonzero_0() -> filter_1(neq_1(|0|_0()))

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:
          
            neq_2#_A(x1,x2) = max{x1, x2}
            s_1_A(x1) = max{23, x1 + 21}
            neq_2#!_A(x1,x2) = max{27, x2 - 22}
            s_1!_A(x1) = 26
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              neq_2#_A(x1,x2) = x1 + x2
              s_1_A(x1) = x1 + 2
              neq_2#!_A(x1,x2) = x1 + x2 + 7
              s_1!_A(x1) = x1 + 6
        
  
    partial status:
  
      pi(neq_2#) = [2]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.