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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{8, x2 + 5}
        app_A(x1,x2) = max{x1 + 4, x2 + 12}
        filter_A = 4
        cons_A = 29
        filter2_A = 50
        false_A = 43
        true_A = 38
        map_A = 58
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = 0
          app_A(x1,x2) = 53
          filter_A = 43
          cons_A = 29
          filter2_A = 0
          false_A = 0
          true_A = 0
          map_A = 43
          app#!_A(x1,x2) = 3
          app!_A(x1,x2) = max{7, x1 - 44, x2 + 4}
          filter!_A = 1
          cons!_A = 5
          filter2!_A = 57
          false!_A = 2
          true!_A = 6
          map!_A = 10
  
    partial status:
  
      pi(app#) = [2]
      pi(app) = [1, 2]
      pi(filter) = []
      pi(cons) = []
      pi(filter2) = []
      pi(false) = []
      pi(true) = []
      pi(map) = []

The following rules are removed:

  p1, p3, p5, 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(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(|:|(),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:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{1, x1}
        app_A(x1,x2) = max{4, x1, x2}
        |:|_A = 5
        +_A = 0
        map_A = 5
        nil_A = 1
        filter_A = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = 0
          app_A(x1,x2) = max{34, x1 + 4, x2 + 15}
          |:|_A = 7
          +_A = 26
          map_A = 45
          nil_A = 80
          filter_A = 11
          app#!_A(x1,x2) = 1
          app!_A(x1,x2) = max{2, x1 - 4, x2 - 34}
          |:|!_A = 51
          +!_A = 50
          map!_A = 58
          nil!_A = 46
          filter!_A = 61
  
    partial status:
  
      pi(app#) = [1]
      pi(app) = []
      pi(|:|) = []
      pi(+) = []
      pi(map) = []
      pi(nil) = []
      pi(filter) = []

The following rules are removed:

  p1, p2, p3, p4

Then no dependency pair remains.