YES
We show the termination of the TRS R:

  app(not(),app(not(),x)) -> x
  app(not(),app(app(or(),x),y)) -> app(app(and(),app(not(),x)),app(not(),y))
  app(not(),app(app(and(),x),y)) -> app(app(or(),app(not(),x)),app(not(),y))
  app(app(and(),x),app(app(or(),y),z)) -> app(app(or(),app(app(and(),x),y)),app(app(and(),x),z))
  app(app(and(),app(app(or(),y),z)),x) -> app(app(or(),app(app(and(),x),y)),app(app(and(),x),z))
  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)

Instead we prove termination of the uncurried system:

  app(and_0(),y) -> and_1(y)
  app(and_1(x1),y) -> and_2(x1,y)
  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(filter_0(),y) -> filter_1(y)
  app(filter_1(x1),y) -> filter_2(x1,y)
  app(filter2_0(),y) -> filter2_1(y)
  app(filter2_1(x1),y) -> filter2_2(x1,y)
  app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
  app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(not_0(),y) -> not_1(y)
  app(or_0(),y) -> or_1(y)
  app(or_1(x1),y) -> or_2(x1,y)
  not_1(not_1(x)) -> x
  not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y))
  not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y))
  and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z))
  and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z))
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  filter_2(f,nil_0()) -> nil_0()
  filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
  filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
  filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)


-- SCC decomposition.

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

p1: app#(and_1(x1),y) -> and_2#(x1,y)
p2: app#(filter_1(x1),y) -> filter_2#(x1,y)
p3: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p4: app#(map_1(x1),y) -> map_2#(x1,y)
p5: app#(not_0(),y) -> not_1#(y)
p6: not_1#(or_2(x,y)) -> and_2#(not_1(x),not_1(y))
p7: not_1#(or_2(x,y)) -> not_1#(x)
p8: not_1#(or_2(x,y)) -> not_1#(y)
p9: not_1#(and_2(x,y)) -> not_1#(x)
p10: not_1#(and_2(x,y)) -> not_1#(y)
p11: and_2#(x,or_2(y,z)) -> and_2#(x,y)
p12: and_2#(x,or_2(y,z)) -> and_2#(x,z)
p13: and_2#(or_2(y,z),x) -> and_2#(x,y)
p14: and_2#(or_2(y,z),x) -> and_2#(x,z)
p15: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p16: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p17: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p18: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p19: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p20: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(and_0(),y) -> and_1(y)
r2: app(and_1(x1),y) -> and_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(not_0(),y) -> not_1(y)
r14: app(or_0(),y) -> or_1(y)
r15: app(or_1(x1),y) -> or_2(x1,y)
r16: not_1(not_1(x)) -> x
r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y))
r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y))
r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z))
r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p15, p16, p17, p18, p19, p20}
  {p7, p8, p9, p10}
  {p11, p12, p13, p14}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p4: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p5: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p6: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p7: app#(filter_1(x1),y) -> filter_2#(x1,y)
p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p9: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(and_0(),y) -> and_1(y)
r2: app(and_1(x1),y) -> and_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(not_0(),y) -> not_1(y)
r14: app(or_0(),y) -> or_1(y)
r15: app(or_1(x1),y) -> or_2(x1,y)
r16: not_1(not_1(x)) -> x
r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y))
r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y))
r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z))
r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r16

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x1 + x2
              cons_2_A(x1,x2) = x1 + x2 + 6
              app#_A(x1,x2) = x1 + x2
              map_1_A(x1) = x1 + 3
              filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 7
              filter2_4#_A(x1,x2,x3,x4) = x2 + x4
              false_0_A() = 1
              filter_2#_A(x1,x2) = x1 + x2
              filter_1_A(x1) = x1
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              not_1_A(x1) = x1 + 3
              map_2#!_A(x1,x2) = 4
              cons_2!_A(x1,x2) = x1 + x2 + 8
              app#!_A(x1,x2) = x2 + 4
              map_1!_A(x1) = x1 + 5
              filter2_3!_A(x1,x2,x3) = 0
              filter2_4#!_A(x1,x2,x3,x4) = x4 + 3
              false_0!_A() = 3
              filter_2#!_A(x1,x2) = x2 + 2
              filter_1!_A(x1) = 1
              app!_A(x1,x2) = 7
              true_0!_A() = 21
              not_1!_A(x1) = 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 1, x2 - 8}
            cons_2_A(x1,x2) = max{x1 + 7, x2}
            app#_A(x1,x2) = max{1, x1 - 33, x2 - 1}
            map_1_A(x1) = x1 + 32
            filter2_3_A(x1,x2,x3) = max{x1 + 32, x2 + 32, x3 + 32}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 33, x3 - 1, x4 - 8}
            false_0_A = 0
            filter_2#_A(x1,x2) = max{1, x1 - 33, x2 - 8}
            filter_1_A(x1) = x1
            app_A(x1,x2) = max{1, x1 - 48, x2 - 48}
            true_0_A = 2
            not_1_A(x1) = max{46, x1 + 7}
            map_2#!_A(x1,x2) = 4
            cons_2!_A(x1,x2) = max{x1 + 5, x2 + 50}
            app#!_A(x1,x2) = max{4, x2 + 2}
            map_1!_A(x1) = max{7, x1 - 3}
            filter2_3!_A(x1,x2,x3) = max{8, x1 + 1, x2 - 3, x3 - 2}
            filter2_4#!_A(x1,x2,x3,x4) = max{9, x3 - 1, x4 + 1}
            false_0!_A = 1
            filter_2#!_A(x1,x2) = max{3, x2}
            filter_1!_A(x1) = max{2, x1 - 4}
            app!_A(x1,x2) = max{6, x2 - 5}
            true_0!_A = 54
            not_1!_A(x1) = max{39, x1 + 25}
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app#) = [2]
      pi(map_1) = []
      pi(filter2_3) = []
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(filter_1) = []
      pi(app) = [2]
      pi(true_0) = []
      pi(not_1) = [1]

The following rules are removed:

  p1, p3, p4, p5, p6, p7, p8, p9



-- SCC decomposition.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)

and R consists of:

r1: app(and_0(),y) -> and_1(y)
r2: app(and_1(x1),y) -> and_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(not_0(),y) -> not_1(y)
r14: app(or_0(),y) -> or_1(y)
r15: app(or_1(x1),y) -> or_2(x1,y)
r16: not_1(not_1(x)) -> x
r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y))
r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y))
r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z))
r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(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: not_1#(and_2(x,y)) -> not_1#(y)
p2: not_1#(and_2(x,y)) -> not_1#(x)
p3: not_1#(or_2(x,y)) -> not_1#(y)
p4: not_1#(or_2(x,y)) -> not_1#(x)

and R consists of:

r1: app(and_0(),y) -> and_1(y)
r2: app(and_1(x1),y) -> and_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(not_0(),y) -> not_1(y)
r14: app(or_0(),y) -> or_1(y)
r15: app(or_1(x1),y) -> or_2(x1,y)
r16: not_1(not_1(x)) -> x
r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y))
r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y))
r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z))
r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              not_1#_A(x1) = x1
              and_2_A(x1,x2) = x1 + x2 + 5
              or_2_A(x1,x2) = x1 + x2 + 5
              not_1#!_A(x1) = x1 + 2
              and_2!_A(x1,x2) = x1 + 2
              or_2!_A(x1,x2) = 8
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            not_1#_A(x1) = 0
            and_2_A(x1,x2) = max{15, x1 - 22, x2 - 22}
            or_2_A(x1,x2) = max{2, x1 - 22, x2 - 22}
            not_1#!_A(x1) = max{17, x1 - 20}
            and_2!_A(x1,x2) = max{x1 + 17, x2 - 3}
            or_2!_A(x1,x2) = max{3, x1 - 19, x2 - 10}
        
  
    partial status:
  
      pi(not_1#) = []
      pi(and_2) = [2]
      pi(or_2) = [1, 2]

The following rules are removed:

  p1, p2, p3, p4

Then no dependency pair remains.

-- Reduction pair.

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

p1: and_2#(x,or_2(y,z)) -> and_2#(x,y)
p2: and_2#(or_2(y,z),x) -> and_2#(x,z)
p3: and_2#(or_2(y,z),x) -> and_2#(x,y)
p4: and_2#(x,or_2(y,z)) -> and_2#(x,z)

and R consists of:

r1: app(and_0(),y) -> and_1(y)
r2: app(and_1(x1),y) -> and_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(not_0(),y) -> not_1(y)
r14: app(or_0(),y) -> or_1(y)
r15: app(or_1(x1),y) -> or_2(x1,y)
r16: not_1(not_1(x)) -> x
r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y))
r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y))
r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z))
r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              and_2#_A(x1,x2) = x1 + x2
              or_2_A(x1,x2) = x1 + x2 + 5
              and_2#!_A(x1,x2) = 0
              or_2!_A(x1,x2) = x2 + 11
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            and_2#_A(x1,x2) = 0
            or_2_A(x1,x2) = max{8, x1 + 4, x2 + 3}
            and_2#!_A(x1,x2) = max{x1 - 1, x2 + 2}
            or_2!_A(x1,x2) = 16
        
  
    partial status:
  
      pi(and_2#) = []
      pi(or_2) = []

The following rules are removed:

  p3



-- SCC decomposition.

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

p1: and_2#(x,or_2(y,z)) -> and_2#(x,y)
p2: and_2#(or_2(y,z),x) -> and_2#(x,z)
p3: and_2#(x,or_2(y,z)) -> and_2#(x,z)

and R consists of:

r1: app(and_0(),y) -> and_1(y)
r2: app(and_1(x1),y) -> and_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(not_0(),y) -> not_1(y)
r14: app(or_0(),y) -> or_1(y)
r15: app(or_1(x1),y) -> or_2(x1,y)
r16: not_1(not_1(x)) -> x
r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y))
r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y))
r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z))
r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

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: and_2#(x,or_2(y,z)) -> and_2#(x,y)
p2: and_2#(x,or_2(y,z)) -> and_2#(x,z)
p3: and_2#(or_2(y,z),x) -> and_2#(x,z)

and R consists of:

r1: app(and_0(),y) -> and_1(y)
r2: app(and_1(x1),y) -> and_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(not_0(),y) -> not_1(y)
r14: app(or_0(),y) -> or_1(y)
r15: app(or_1(x1),y) -> or_2(x1,y)
r16: not_1(not_1(x)) -> x
r17: not_1(or_2(x,y)) -> and_2(not_1(x),not_1(y))
r18: not_1(and_2(x,y)) -> or_2(not_1(x),not_1(y))
r19: and_2(x,or_2(y,z)) -> or_2(and_2(x,y),and_2(x,z))
r20: and_2(or_2(y,z),x) -> or_2(and_2(x,y),and_2(x,z))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              and_2#_A(x1,x2) = 0
              or_2_A(x1,x2) = x1 + x2 + 5
              and_2#!_A(x1,x2) = x1 + x2 + 6
              or_2!_A(x1,x2) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            and_2#_A(x1,x2) = 0
            or_2_A(x1,x2) = max{15, x1 - 16, x2 - 16}
            and_2#!_A(x1,x2) = max{30, x2 + 15}
            or_2!_A(x1,x2) = x1 + 28
        
  
    partial status:
  
      pi(and_2#) = []
      pi(or_2) = [2]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.