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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = x1 + 10
          app_A(x1,x2) = x2 + 35
          filter_A = 8
          cons_A = 21
          filtersub_A = 1
          false_A = 10
          true_A = 40
          neq_A = 11
          |0|_A = 6
          s_A = 0
          nil_A = 9
    
      precedence:
      
        app = filtersub = false = true = s > app# > filter > cons > nil > neq = |0|
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(filter) = []
        pi(cons) = []
        pi(filtersub) = []
        pi(false) = []
        pi(true) = []
        pi(neq) = []
        pi(|0|) = []
        pi(s) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = 0
          app_A(x1,x2) = 48
          filter_A = 72
          cons_A = 69
          filtersub_A = 66
          false_A = 68
          true_A = 68
          neq_A = 67
          |0|_A = 0
          s_A = 66
          nil_A = 73
    
      precedence:
      
        nil > filter = cons > filtersub = false = true > neq > app > app# > |0| = s
      
      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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 - 21, x2 + 48}
          app_A(x1,x2) = x2 + 37
          filtersub_A = 35
          false_A = 5
          cons_A = 21
          filter_A = 33
          true_A = 45
          neq_A = 37
          |0|_A = 46
          s_A = 46
          nil_A = 28
    
      precedence:
      
        neq > filtersub = |0| = s > cons = nil > app# = app = false = filter = true
      
      partial status:
    
        pi(app#) = [2]
        pi(app) = []
        pi(filtersub) = []
        pi(false) = []
        pi(cons) = []
        pi(filter) = []
        pi(true) = []
        pi(neq) = []
        pi(|0|) = []
        pi(s) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = 17
          app_A(x1,x2) = 17
          filtersub_A = 1
          false_A = 12
          cons_A = 12
          filter_A = 16
          true_A = 8
          neq_A = 18
          |0|_A = 0
          s_A = 4
          nil_A = 18
    
      precedence:
      
        filtersub > app# = cons = |0| > app > false > nil > filter = true = neq = s
      
      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:

  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))
p2: 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}


-- Reduction pair.

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))
p2: 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = x2 + 4
          app_A(x1,x2) = x2 + 60
          filter_A = 31
          cons_A = 53
          filtersub_A = 35
          true_A = 0
          neq_A = 47
          |0|_A = 33
          false_A = 32
          s_A = 43
          nil_A = 30
    
      precedence:
      
        app = neq > cons > filtersub > app# > true > |0| = false > s = nil > filter
      
      partial status:
    
        pi(app#) = [2]
        pi(app) = []
        pi(filter) = []
        pi(cons) = []
        pi(filtersub) = []
        pi(true) = []
        pi(neq) = []
        pi(|0|) = []
        pi(false) = []
        pi(s) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{0, x2 - 28}
          app_A(x1,x2) = 29
          filter_A = 50
          cons_A = 47
          filtersub_A = 44
          true_A = 28
          neq_A = 30
          |0|_A = 0
          false_A = 47
          s_A = 27
          nil_A = 51
    
      precedence:
      
        app# = app = filter = cons = filtersub = true = neq = |0| = false = s = nil
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(filter) = []
        pi(cons) = []
        pi(filtersub) = []
        pi(true) = []
        pi(neq) = []
        pi(|0|) = []
        pi(false) = []
        pi(s) = []
        pi(nil) = []
    

The next rules are strictly ordered:

  p2

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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 + 2, x2 + 4}
          app_A(x1,x2) = max{x1, x2 + 3}
          neq_A = 2
          s_A = 3
    
      precedence:
      
        app# = app = neq = s
      
      partial status:
    
        pi(app#) = [1, 2]
        pi(app) = [1]
        pi(neq) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 + 2, x2 + 4}
          app_A(x1,x2) = max{4, x1}
          neq_A = 5
          s_A = 6
    
      precedence:
      
        app# = app = neq = s
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1]
        pi(neq) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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