YES
We show the termination of the TRS R:

  app(app(|:|(),app(app(|:|(),x),y)),z) -> app(app(|:|(),x),app(app(|:|(),y),z))
  app(app(|:|(),app(app(+(),x),y)),z) -> app(app(+(),app(app(|:|(),x),z)),app(app(|:|(),y),z))
  app(app(|:|(),z),app(app(+(),x),app(f,y))) -> app(app(|:|(),app(app(g(),z),y)),app(app(+(),x),a()))
  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
  app(app(filter(),f),nil()) -> nil()
  app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs)
  app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs))
  app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs)


-- SCC decomposition.

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

p1: app#(app(|:|(),app(app(|:|(),x),y)),z) -> app#(app(|:|(),x),app(app(|:|(),y),z))
p2: app#(app(|:|(),app(app(|:|(),x),y)),z) -> app#(app(|:|(),y),z)
p3: app#(app(|:|(),app(app(|:|(),x),y)),z) -> app#(|:|(),y)
p4: app#(app(|:|(),app(app(+(),x),y)),z) -> app#(app(+(),app(app(|:|(),x),z)),app(app(|:|(),y),z))
p5: app#(app(|:|(),app(app(+(),x),y)),z) -> app#(+(),app(app(|:|(),x),z))
p6: app#(app(|:|(),app(app(+(),x),y)),z) -> app#(app(|:|(),x),z)
p7: app#(app(|:|(),app(app(+(),x),y)),z) -> app#(|:|(),x)
p8: app#(app(|:|(),app(app(+(),x),y)),z) -> app#(app(|:|(),y),z)
p9: app#(app(|:|(),app(app(+(),x),y)),z) -> app#(|:|(),y)
p10: app#(app(|:|(),z),app(app(+(),x),app(f,y))) -> app#(app(|:|(),app(app(g(),z),y)),app(app(+(),x),a()))
p11: app#(app(|:|(),z),app(app(+(),x),app(f,y))) -> app#(|:|(),app(app(g(),z),y))
p12: app#(app(|:|(),z),app(app(+(),x),app(f,y))) -> app#(app(g(),z),y)
p13: app#(app(|:|(),z),app(app(+(),x),app(f,y))) -> app#(g(),z)
p14: app#(app(|:|(),z),app(app(+(),x),app(f,y))) -> app#(app(+(),x),a())
p15: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(cons(),app(f,x)),app(app(map(),f),xs))
p16: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(cons(),app(f,x))
p17: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p18: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)
p19: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(app(app(filter2(),app(f,x)),f),x),xs)
p20: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(app(filter2(),app(f,x)),f),x)
p21: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(filter2(),app(f,x)),f)
p22: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(filter2(),app(f,x))
p23: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(f,x)
p24: app#(app(app(app(filter2(),true()),f),x),xs) -> app#(app(cons(),x),app(app(filter(),f),xs))
p25: app#(app(app(app(filter2(),true()),f),x),xs) -> app#(cons(),x)
p26: app#(app(app(app(filter2(),true()),f),x),xs) -> app#(app(filter(),f),xs)
p27: app#(app(app(app(filter2(),true()),f),x),xs) -> app#(filter(),f)
p28: app#(app(app(app(filter2(),false()),f),x),xs) -> app#(app(filter(),f),xs)
p29: app#(app(app(app(filter2(),false()),f),x),xs) -> app#(filter(),f)

and R consists of:

r1: app(app(|:|(),app(app(|:|(),x),y)),z) -> app(app(|:|(),x),app(app(|:|(),y),z))
r2: app(app(|:|(),app(app(+(),x),y)),z) -> app(app(+(),app(app(|:|(),x),z)),app(app(|:|(),y),z))
r3: app(app(|:|(),z),app(app(+(),x),app(f,y))) -> app(app(|:|(),app(app(g(),z),y)),app(app(+(),x),a()))
r4: app(app(map(),f),nil()) -> nil()
r5: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r6: app(app(filter(),f),nil()) -> nil()
r7: app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs)
r8: app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs))
r9: app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs)

The estimated dependency graph contains the following SCCs:

  {p17, p18, p19, p23, p26, p28}
  {p1, p2, p6, p8}


-- Reduction pair.

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

p1: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(f,x)
p2: app#(app(app(app(filter2(),false()),f),x),xs) -> app#(app(filter(),f),xs)
p3: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(app(app(filter2(),app(f,x)),f),x),xs)
p4: app#(app(app(app(filter2(),true()),f),x),xs) -> app#(app(filter(),f),xs)
p5: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)
p6: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)

and R consists of:

r1: app(app(|:|(),app(app(|:|(),x),y)),z) -> app(app(|:|(),x),app(app(|:|(),y),z))
r2: app(app(|:|(),app(app(+(),x),y)),z) -> app(app(+(),app(app(|:|(),x),z)),app(app(|:|(),y),z))
r3: app(app(|:|(),z),app(app(+(),x),app(f,y))) -> app(app(|:|(),app(app(g(),z),y)),app(app(+(),x),a()))
r4: app(app(map(),f),nil()) -> nil()
r5: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r6: app(app(filter(),f),nil()) -> nil()
r7: app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs)
r8: app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs))
r9: app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs)

The set of usable rules consists of

  r4, r6

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{0, x2 - 7}
            app_A(x1,x2) = max{6, x1, x2}
            filter_A = 3
            cons_A = 3
            filter2_A = 5
            false_A = 5
            true_A = 2
            map_A = 7
            nil_A = 8
            app#!_A(x1,x2) = 0
            app!_A(x1,x2) = 3
            filter!_A = 5
            cons!_A = 5
            filter2!_A = 4
            false!_A = 0
            true!_A = 0
            map!_A = 6
            nil!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x2 - 4}
            app_A(x1,x2) = max{24, x1 + 4, x2}
            filter_A = 8
            cons_A = 12
            filter2_A = 7
            false_A = 3
            true_A = 3
            map_A = 22
            nil_A = 31
            app#!_A(x1,x2) = x2 + 6
            app!_A(x1,x2) = 3
            filter!_A = 36
            cons!_A = 36
            filter2!_A = 35
            false!_A = 4
            true!_A = 4
            map!_A = 3
            nil!_A = 3
        
  
    partial status:
  
      pi(app#) = []
      pi(app) = [1]
      pi(filter) = []
      pi(cons) = []
      pi(filter2) = []
      pi(false) = []
      pi(true) = []
      pi(map) = []
      pi(nil) = []

The following rules are removed:

  p1, p6



-- SCC decomposition.

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

p1: app#(app(app(app(filter2(),false()),f),x),xs) -> app#(app(filter(),f),xs)
p2: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(app(app(filter2(),app(f,x)),f),x),xs)
p3: app#(app(app(app(filter2(),true()),f),x),xs) -> app#(app(filter(),f),xs)
p4: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)

and R consists of:

r1: app(app(|:|(),app(app(|:|(),x),y)),z) -> app(app(|:|(),x),app(app(|:|(),y),z))
r2: app(app(|:|(),app(app(+(),x),y)),z) -> app(app(+(),app(app(|:|(),x),z)),app(app(|:|(),y),z))
r3: app(app(|:|(),z),app(app(+(),x),app(f,y))) -> app(app(|:|(),app(app(g(),z),y)),app(app(+(),x),a()))
r4: app(app(map(),f),nil()) -> nil()
r5: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r6: app(app(filter(),f),nil()) -> nil()
r7: app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs)
r8: app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs))
r9: app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}
  {p4}


-- Reduction pair.

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

p1: app#(app(app(app(filter2(),false()),f),x),xs) -> app#(app(filter(),f),xs)
p2: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(app(app(filter2(),app(f,x)),f),x),xs)
p3: app#(app(app(app(filter2(),true()),f),x),xs) -> app#(app(filter(),f),xs)

and R consists of:

r1: app(app(|:|(),app(app(|:|(),x),y)),z) -> app(app(|:|(),x),app(app(|:|(),y),z))
r2: app(app(|:|(),app(app(+(),x),y)),z) -> app(app(+(),app(app(|:|(),x),z)),app(app(|:|(),y),z))
r3: app(app(|:|(),z),app(app(+(),x),app(f,y))) -> app(app(|:|(),app(app(g(),z),y)),app(app(+(),x),a()))
r4: app(app(map(),f),nil()) -> nil()
r5: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r6: app(app(filter(),f),nil()) -> nil()
r7: app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs)
r8: app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs))
r9: app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs)

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:
          
            app#_A(x1,x2) = 0
            app_A(x1,x2) = 26
            filter2_A = 3
            false_A = 1
            filter_A = 35
            cons_A = 13
            true_A = 26
            app#!_A(x1,x2) = 0
            app!_A(x1,x2) = max{x1 + 5, x2 + 5}
            filter2!_A = 6
            false!_A = 0
            filter!_A = 7
            cons!_A = 0
            true!_A = 17
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 0
            app_A(x1,x2) = max{20, x1 + 5, x2 - 5}
            filter2_A = 15
            false_A = 17
            filter_A = 18
            cons_A = 29
            true_A = 21
            app#!_A(x1,x2) = max{65, x2 + 7}
            app!_A(x1,x2) = x1 + 30
            filter2!_A = 44
            false!_A = 44
            filter!_A = 45
            cons!_A = 60
            true!_A = 49
        
  
    partial status:
  
      pi(app#) = [2]
      pi(app) = [2]
      pi(filter2) = []
      pi(false) = []
      pi(filter) = []
      pi(cons) = []
      pi(true) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: app#(app(app(app(filter2(),false()),f),x),xs) -> app#(app(filter(),f),xs)
p2: app#(app(app(app(filter2(),true()),f),x),xs) -> app#(app(filter(),f),xs)

and R consists of:

r1: app(app(|:|(),app(app(|:|(),x),y)),z) -> app(app(|:|(),x),app(app(|:|(),y),z))
r2: app(app(|:|(),app(app(+(),x),y)),z) -> app(app(+(),app(app(|:|(),x),z)),app(app(|:|(),y),z))
r3: app(app(|:|(),z),app(app(+(),x),app(f,y))) -> app(app(|:|(),app(app(g(),z),y)),app(app(+(),x),a()))
r4: app(app(map(),f),nil()) -> nil()
r5: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r6: app(app(filter(),f),nil()) -> nil()
r7: app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs)
r8: app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs))
r9: app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs)

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(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)

and R consists of:

r1: app(app(|:|(),app(app(|:|(),x),y)),z) -> app(app(|:|(),x),app(app(|:|(),y),z))
r2: app(app(|:|(),app(app(+(),x),y)),z) -> app(app(+(),app(app(|:|(),x),z)),app(app(|:|(),y),z))
r3: app(app(|:|(),z),app(app(+(),x),app(f,y))) -> app(app(|:|(),app(app(g(),z),y)),app(app(+(),x),a()))
r4: app(app(map(),f),nil()) -> nil()
r5: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r6: app(app(filter(),f),nil()) -> nil()
r7: app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs)
r8: app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs))
r9: app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs)

The set of usable rules consists of

  r4, r6

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{0, x2 - 4}
            app_A(x1,x2) = max{17, x1 - 29, x2 + 3}
            map_A = 0
            cons_A = 13
            nil_A = 18
            filter_A = 19
            app#!_A(x1,x2) = max{3, x1 - 35, x2 - 18}
            app!_A(x1,x2) = max{3, x1 - 22, x2 - 112}
            map!_A = 2
            cons!_A = 1
            nil!_A = 0
            filter!_A = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x2 - 5}
            app_A(x1,x2) = max{11, x1 - 18, x2 + 4}
            map_A = 27
            cons_A = 6
            nil_A = 12
            filter_A = 31
            app#!_A(x1,x2) = max{3, x1 - 110, x2 - 91}
            app!_A(x1,x2) = max{4, x1 - 106, x2 - 163}
            map!_A = 2
            cons!_A = 1
            nil!_A = 0
            filter!_A = 1
        
  
    partial status:
  
      pi(app#) = [2]
      pi(app) = [2]
      pi(map) = []
      pi(cons) = []
      pi(nil) = []
      pi(filter) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: app#(app(|:|(),app(app(|:|(),x),y)),z) -> app#(app(|:|(),x),app(app(|:|(),y),z))
p2: app#(app(|:|(),app(app(+(),x),y)),z) -> app#(app(|:|(),y),z)
p3: app#(app(|:|(),app(app(+(),x),y)),z) -> app#(app(|:|(),x),z)
p4: app#(app(|:|(),app(app(|:|(),x),y)),z) -> app#(app(|:|(),y),z)

and R consists of:

r1: app(app(|:|(),app(app(|:|(),x),y)),z) -> app(app(|:|(),x),app(app(|:|(),y),z))
r2: app(app(|:|(),app(app(+(),x),y)),z) -> app(app(+(),app(app(|:|(),x),z)),app(app(|:|(),y),z))
r3: app(app(|:|(),z),app(app(+(),x),app(f,y))) -> app(app(|:|(),app(app(g(),z),y)),app(app(+(),x),a()))
r4: app(app(map(),f),nil()) -> nil()
r5: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r6: app(app(filter(),f),nil()) -> nil()
r7: app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs)
r8: app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs))
r9: app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs)

The set of usable rules consists of

  r4, r6

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{0, x1 - 6}
            app_A(x1,x2) = max{x1, x2}
            |:|_A = 0
            +_A = 5
            map_A = 1
            nil_A = 6
            filter_A = 5
            app#!_A(x1,x2) = max{4, x1 - 14}
            app!_A(x1,x2) = max{5, x1 - 10, x2 - 13}
            |:|!_A = 3
            +!_A = 6
            map!_A = 1
            nil!_A = 0
            filter!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 0
            app_A(x1,x2) = max{49, x1 + 45, x2 - 17}
            |:|_A = 4
            +_A = 1
            map_A = 22
            nil_A = 16
            filter_A = 22
            app#!_A(x1,x2) = max{55, x1 - 78}
            app!_A(x1,x2) = max{22, x1 - 72, x2 - 38}
            |:|!_A = 54
            +!_A = 52
            map!_A = 54
            nil!_A = 0
            filter!_A = 66
        
  
    partial status:
  
      pi(app#) = [1]
      pi(app) = [1, 2]
      pi(|:|) = []
      pi(+) = []
      pi(map) = []
      pi(nil) = []
      pi(filter) = []

The following rules are removed:

  p1, p2, p3, p4

Then no dependency pair remains.