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|()))

-- SCC decomposition.

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

p1: app#(app(neq(),app(s(),x)),app(s(),y)) -> app#(app(neq(),x),y)
p2: app#(app(neq(),app(s(),x)),app(s(),y)) -> app#(neq(),x)
p3: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
p4: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(app(filtersub(),app(f,y)),f)
p5: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(filtersub(),app(f,y))
p6: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(f,y)
p7: app#(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app#(app(cons(),y),app(app(filter(),f),ys))
p8: app#(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)
p9: app#(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app#(filter(),f)
p10: app#(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)
p11: app#(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app#(filter(),f)
p12: nonzero#() -> app#(filter(),app(neq(),|0|()))
p13: nonzero#() -> app#(neq(),|0|())

and R consists of:

r1: app(app(neq(),|0|()),|0|()) -> false()
r2: app(app(neq(),|0|()),app(s(),y)) -> true()
r3: app(app(neq(),app(s(),x)),|0|()) -> true()
r4: app(app(neq(),app(s(),x)),app(s(),y)) -> app(app(neq(),x),y)
r5: app(app(filter(),f),nil()) -> nil()
r6: app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
r7: app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys))
r8: app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys)
r9: nonzero() -> app(filter(),app(neq(),|0|()))

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(f,y)
p2: app#(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)
p3: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
p4: app#(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)

and R consists of:

r1: app(app(neq(),|0|()),|0|()) -> false()
r2: app(app(neq(),|0|()),app(s(),y)) -> true()
r3: app(app(neq(),app(s(),x)),|0|()) -> true()
r4: app(app(neq(),app(s(),x)),app(s(),y)) -> app(app(neq(),x),y)
r5: app(app(filter(),f),nil()) -> nil()
r6: app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
r7: app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys))
r8: app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys)
r9: nonzero() -> app(filter(),app(neq(),|0|()))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1
      app_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,0,0),(1,0,0,0)) x2 + (1,1,1,1)
      filter_A() = (4,1,1,1)
      cons_A() = (1,1,1,0)
      filtersub_A() = (1,1,1,0)
      false_A() = (1,3,3,1)
      true_A() = (1,1,1,1)
      neq_A() = (1,1,1,0)
      |0|_A() = (1,1,1,1)
      s_A() = (1,1,1,1)
      nil_A() = (1,1,1,1)

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: app#(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)
p2: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
p3: app#(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)

and R consists of:

r1: app(app(neq(),|0|()),|0|()) -> false()
r2: app(app(neq(),|0|()),app(s(),y)) -> true()
r3: app(app(neq(),app(s(),x)),|0|()) -> true()
r4: app(app(neq(),app(s(),x)),app(s(),y)) -> app(app(neq(),x),y)
r5: app(app(filter(),f),nil()) -> nil()
r6: app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
r7: app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys))
r8: app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys)
r9: nonzero() -> app(filter(),app(neq(),|0|()))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: app#(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)
p2: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
p3: app#(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)

and R consists of:

r1: app(app(neq(),|0|()),|0|()) -> false()
r2: app(app(neq(),|0|()),app(s(),y)) -> true()
r3: app(app(neq(),app(s(),x)),|0|()) -> true()
r4: app(app(neq(),app(s(),x)),app(s(),y)) -> app(app(neq(),x),y)
r5: app(app(filter(),f),nil()) -> nil()
r6: app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
r7: app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys))
r8: app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys)
r9: nonzero() -> app(filter(),app(neq(),|0|()))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x2
      app_A(x1,x2) = x2 + (1,1,1,1)
      filtersub_A() = (1,1,0,0)
      false_A() = (1,2,2,2)
      cons_A() = (1,0,0,0)
      filter_A() = (1,0,0,1)
      true_A() = (1,2,2,2)
      neq_A() = (1,0,0,0)
      |0|_A() = (1,1,1,1)
      s_A() = (1,0,0,1)
      nil_A() = (1,2,2,2)

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))

and R consists of:

r1: app(app(neq(),|0|()),|0|()) -> false()
r2: app(app(neq(),|0|()),app(s(),y)) -> true()
r3: app(app(neq(),app(s(),x)),|0|()) -> true()
r4: app(app(neq(),app(s(),x)),app(s(),y)) -> app(app(neq(),x),y)
r5: app(app(filter(),f),nil()) -> nil()
r6: app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
r7: app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys))
r8: app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys)
r9: nonzero() -> app(filter(),app(neq(),|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: app#(app(neq(),app(s(),x)),app(s(),y)) -> app#(app(neq(),x),y)

and R consists of:

r1: app(app(neq(),|0|()),|0|()) -> false()
r2: app(app(neq(),|0|()),app(s(),y)) -> true()
r3: app(app(neq(),app(s(),x)),|0|()) -> true()
r4: app(app(neq(),app(s(),x)),app(s(),y)) -> app(app(neq(),x),y)
r5: app(app(filter(),f),nil()) -> nil()
r6: app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
r7: app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys))
r8: app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys)
r9: nonzero() -> app(filter(),app(neq(),|0|()))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,0,1,1)) x2
      app_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,0,1,1)) x2 + (1,0,0,0)
      neq_A() = (1,1,1,1)
      s_A() = (1,1,1,1)

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.