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

  r3, 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) = 1
            filter_1_A(x1) = max{8, x1}
            filter_2#_A(x1,x2) = 1
            cons_2_A(x1,x2) = max{25, x1 + 4, x2}
            filtersub_2_A(x1,x2) = max{16, x1 - 25, x2 - 9}
            filtersub_3#_A(x1,x2,x3) = 1
            false_0_A = 25
            app_A(x1,x2) = max{23, x1 + 7, x2}
            true_0_A = 23
            filter_0_A = 0
            filter_2_A(x1,x2) = x2 + 1
            nil_0_A = 30
            app#!_A(x1,x2) = max{24, x2 + 3}
            filter_1!_A(x1) = 2
            filter_2#!_A(x1,x2) = max{0, x2 - 1}
            cons_2!_A(x1,x2) = 27
            filtersub_2!_A(x1,x2) = max{40, x1 + 29}
            filtersub_3#!_A(x1,x2,x3) = max{1, x3 - 1}
            false_0!_A = 0
            app!_A(x1,x2) = max{x1 + 31, x2 + 5}
            true_0!_A = 28
            filter_0!_A = 3
            filter_2!_A(x1,x2) = max{29, x1 + 12, x2 + 1}
            nil_0!_A = 30
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{2, x2 - 1}
            filter_1_A(x1) = 5
            filter_2#_A(x1,x2) = max{1, x2 - 21}
            cons_2_A(x1,x2) = max{23, x1 + 20, x2 + 18}
            filtersub_2_A(x1,x2) = max{6, x1 + 4, x2 + 4}
            filtersub_3#_A(x1,x2,x3) = max{2, x3 - 21}
            false_0_A = 27
            app_A(x1,x2) = 5
            true_0_A = 31
            filter_0_A = 0
            filter_2_A(x1,x2) = 7
            nil_0_A = 0
            app#!_A(x1,x2) = max{33, x2 + 20}
            filter_1!_A(x1) = 30
            filter_2#!_A(x1,x2) = max{21, x2 + 16}
            cons_2!_A(x1,x2) = x2 + 18
            filtersub_2!_A(x1,x2) = max{31, x1 - 1}
            filtersub_3#!_A(x1,x2,x3) = max{32, x3 - 1}
            false_0!_A = 20
            app!_A(x1,x2) = max{30, x1 - 3, x2 - 2}
            true_0!_A = 26
            filter_0!_A = 31
            filter_2!_A(x1,x2) = 17
            nil_0!_A = 1
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = []
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(filtersub_2) = [1]
      pi(filtersub_3#) = []
      pi(false_0) = []
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(filter_0) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3, p4, p5, p6

Then no dependency pair remains.

-- 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{1, x1}
            neq_2#!_A(x1,x2) = max{5, x1 - 19, x2 - 2}
            s_1!_A(x1) = max{4, x1 - 18}
        
        2. 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{1, x1}
            neq_2#!_A(x1,x2) = max{7, x1 - 6, x2 - 7}
            s_1!_A(x1) = max{6, x1 - 6}
        
  
    partial status:
  
      pi(neq_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.