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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((0,1),(0,0)) x1 + (5,9)
          app_A(x1,x2) = ((0,0),(0,1)) x2 + (3,3)
          filter_A() = (4,10)
          cons_A() = (1,8)
          filtersub_A() = (2,2)
          false_A() = (2,2)
          true_A() = (2,4)
          neq_A() = (4,7)
          |0|_A() = (1,1)
          s_A() = (1,0)
          nil_A() = (1,11)
  
    precedence:
    
      app# = cons = filtersub = false > app = filter = true = neq = |0| = s = nil
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(filter) = []
      pi(cons) = []
      pi(filtersub) = []
      pi(false) = []
      pi(true) = []
      pi(neq) = []
      pi(|0|) = []
      pi(s) = []
      pi(nil) = []

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((0,1),(0,0)) x2 + (4,31)
          app_A(x1,x2) = ((0,0),(1,1)) x2 + (10,3)
          filtersub_A() = (11,30)
          false_A() = (9,18)
          cons_A() = (9,17)
          filter_A() = (8,18)
          true_A() = (3,2)
          neq_A() = (2,16)
          |0|_A() = (2,13)
          s_A() = (1,1)
          nil_A() = (9,19)
  
    precedence:
    
      app# = app = filtersub = false = filter = true = neq > cons = |0| = s = nil
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(filtersub) = []
      pi(false) = []
      pi(cons) = []
      pi(filter) = []
      pi(true) = []
      pi(neq) = []
      pi(|0|) = []
      pi(s) = []
      pi(nil) = []

The next rules are strictly ordered:

  p3

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))

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}


-- 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))

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((0,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (2,0)
          app_A(x1,x2) = ((1,0),(0,0)) x2 + (3,1)
          filtersub_A() = (7,8)
          false_A() = (0,1)
          cons_A() = (1,3)
          filter_A() = (2,6)
          neq_A() = (7,2)
          |0|_A() = (1,1)
          s_A() = (1,1)
          true_A() = (0,0)
          nil_A() = (1,1)
  
    precedence:
    
      app# = filtersub = false > filter = neq = |0| = s = true = nil > app = cons
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(filtersub) = []
      pi(false) = []
      pi(cons) = []
      pi(filter) = []
      pi(neq) = []
      pi(|0|) = []
      pi(s) = []
      pi(true) = []
      pi(nil) = []

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(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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          app#_A(x1,x2) = ((1,1),(1,1)) x2 + (7,5)
          app_A(x1,x2) = x2 + (3,2)
          neq_A() = (1,1)
          s_A() = (2,1)
  
    precedence:
    
      app# = app = s > neq
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(neq) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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