YES
We show the termination of the TRS R:

  app(app(ack(),|0|()),y) -> app(succ(),y)
  app(app(ack(),app(succ(),x)),y) -> app(app(ack(),x),app(succ(),|0|()))
  app(app(ack(),app(succ(),x)),app(succ(),y)) -> app(app(ack(),x),app(app(ack(),app(succ(),x)),y))
  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(ack_0(),y) -> ack_1(y)
  app(ack_1(x1),y) -> ack_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(succ_0(),y) -> succ_1(y)
  ack_2(|0|_0(),y) -> succ_1(y)
  ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
  ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
  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#(ack_1(x1),y) -> ack_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: ack_2#(succ_1(x),y) -> ack_2#(x,succ_1(|0|_0()))
p6: ack_2#(succ_1(x),succ_1(y)) -> ack_2#(x,ack_2(succ_1(x),y))
p7: ack_2#(succ_1(x),succ_1(y)) -> ack_2#(succ_1(x),y)
p8: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p9: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p10: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p11: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p12: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p13: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_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(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p8, p9, p10, p11, p12, p13}
  {p5, p6, p7}


-- 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(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_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(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r3, r7, r13, r17, r19

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:
          
            map_2#_A(x1,x2) = max{38, x1 - 1, x2 - 45}
            cons_2_A(x1,x2) = max{44, x1 + 21, x2 + 21}
            app#_A(x1,x2) = max{38, x1 - 26, x2 - 24}
            map_1_A(x1) = max{29, x1 + 25}
            filter2_3_A(x1,x2,x3) = max{x1 + 38, x2 + 39, x3 + 63}
            filter2_4#_A(x1,x2,x3,x4) = max{38, x2 - 1, x4 - 24}
            false_0_A = 44
            filter_2#_A(x1,x2) = max{38, x1 - 1, x2 - 24}
            filter_1_A(x1) = x1 + 41
            app_A(x1,x2) = max{44, x1 - 42, x2 + 21}
            true_0_A = 44
            ack_0_A = 22
            ack_1_A(x1) = max{35, x1 + 21}
            cons_0_A = 22
            cons_1_A(x1) = max{22, x1 + 21}
            filter2_0_A = 22
            filter2_1_A(x1) = max{22, x1 + 21}
            succ_0_A = 22
            succ_1_A(x1) = max{22, x1}
            map_2_A(x1,x2) = max{22, x1 - 17, x2 + 21}
            nil_0_A = 23
            filter_2_A(x1,x2) = max{23, x1 - 17, x2 - 17}
            map_2#!_A(x1,x2) = max{2, x1 - 4, x2 - 1}
            cons_2!_A(x1,x2) = x1 + 8
            app#!_A(x1,x2) = max{x1 - 28, x2 + 19}
            map_1!_A(x1) = 18
            filter2_3!_A(x1,x2,x3) = max{x1 + 4, x2 + 4, x3 + 6}
            filter2_4#!_A(x1,x2,x3,x4) = max{x2 + 11, x3 + 9, x4 + 18}
            false_0!_A = 1
            filter_2#!_A(x1,x2) = max{12, x1 + 11, x2 - 1}
            filter_1!_A(x1) = max{13, x1 - 1}
            app!_A(x1,x2) = max{x1 - 23, x2 + 17}
            true_0!_A = 1
            ack_0!_A = 1
            ack_1!_A(x1) = max{10, x1 - 23}
            cons_0!_A = 1
            cons_1!_A(x1) = max{0, x1 - 1}
            filter2_0!_A = 1
            filter2_1!_A(x1) = max{2, x1 - 4}
            succ_0!_A = 1
            succ_1!_A(x1) = max{2, x1 - 25}
            map_2!_A(x1,x2) = 7
            nil_0!_A = 0
            filter_2!_A(x1,x2) = max{0, x1 - 1, x2 - 45}
        
        2. 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) = 0
              cons_2_A(x1,x2) = 1
              app#_A(x1,x2) = 0
              map_1_A(x1) = x1 + 3
              filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 3
              filter2_4#_A(x1,x2,x3,x4) = 0
              false_0_A() = 1
              filter_2#_A(x1,x2) = 0
              filter_1_A(x1) = x1 + 3
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              ack_0_A() = 1
              ack_1_A(x1) = x1 + 1
              cons_0_A() = 1
              cons_1_A(x1) = x1 + 1
              filter2_0_A() = 1
              filter2_1_A(x1) = x1 + 1
              succ_0_A() = 3
              succ_1_A(x1) = x1
              map_2_A(x1,x2) = 1
              nil_0_A() = 1
              filter_2_A(x1,x2) = x1 + 1
              map_2#!_A(x1,x2) = x1 + 6
              cons_2!_A(x1,x2) = x1 + 5
              app#!_A(x1,x2) = x1 + 4
              map_1!_A(x1) = 2
              filter2_3!_A(x1,x2,x3) = x2 + x3 + 5
              filter2_4#!_A(x1,x2,x3,x4) = 6
              false_0!_A() = 0
              filter_2#!_A(x1,x2) = 6
              filter_1!_A(x1) = 7
              app!_A(x1,x2) = x1 + 2
              true_0!_A() = 0
              ack_0!_A() = 0
              ack_1!_A(x1) = 1
              cons_0!_A() = 0
              cons_1!_A(x1) = 1
              filter2_0!_A() = 0
              filter2_1!_A(x1) = 1
              succ_0!_A() = 0
              succ_1!_A(x1) = 4
              map_2!_A(x1,x2) = 1
              nil_0!_A() = 0
              filter_2!_A(x1,x2) = 1
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app#) = []
      pi(map_1) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(filter_1) = []
      pi(app) = []
      pi(true_0) = []
      pi(ack_0) = []
      pi(ack_1) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(succ_0) = []
      pi(succ_1) = []
      pi(map_2) = []
      pi(nil_0) = []
      pi(filter_2) = []

The following rules are removed:

  p1, p2, p4, p6, p7



-- SCC decomposition.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p2: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p3: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p4: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_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(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: filter2_4(false_0(),f,x,xs) -> filter_2(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: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_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(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: 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!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{11, x1 - 6, x2 - 7}
            map_2#!_A(x1,x2) = 1
            cons_2!_A(x1,x2) = max{3, x1 - 45, x2 - 21}
        
        2. 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) = x2
              cons_2_A(x1,x2) = x1 + x2 + 5
              map_2#!_A(x1,x2) = x2 + 2
              cons_2!_A(x1,x2) = x1 + x2 + 3
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []

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: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p2: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p3: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_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(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r8, r11, r13, r14, r19

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:
          
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 1, x3 - 103, x4 - 17}
            false_0_A = 111
            filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 25}
            cons_2_A(x1,x2) = max{24, x1 - 78, x2 + 8}
            app_A(x1,x2) = max{77, x1 - 20, x2 + 8}
            true_0_A = 51
            filter2_1_A(x1) = max{9, x1 + 8}
            filter2_2_A(x1,x2) = max{9, x1 - 12, x2 - 24}
            map_0_A = 9
            map_1_A(x1) = max{9, x1 + 8}
            succ_0_A = 9
            succ_1_A(x1) = max{0, x1 - 13}
            ack_2_A(x1,x2) = max{9, x1 - 12, x2 - 10}
            |0|_0_A = 13
            filter_2_A(x1,x2) = max{60, x1 - 72, x2}
            nil_0_A = 61
            filter2_4#!_A(x1,x2,x3,x4) = max{0, x3 - 1}
            false_0!_A = 0
            filter_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = max{9, x1 + 6, x2 + 6}
            app!_A(x1,x2) = max{x1 - 10, x2 + 7}
            true_0!_A = 0
            filter2_1!_A(x1) = max{7, x1 - 135}
            filter2_2!_A(x1,x2) = max{7, x1 - 134, x2 - 155}
            map_0!_A = 1
            map_1!_A(x1) = max{0, x1 - 1}
            succ_0!_A = 1
            succ_1!_A(x1) = x1 + 4
            ack_2!_A(x1,x2) = x2 + 5
            |0|_0!_A = 2
            filter_2!_A(x1,x2) = max{10, x1 - 153, x2 - 62}
            nil_0!_A = 9
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              filter2_4#_A(x1,x2,x3,x4) = x4
              false_0_A() = 1
              filter_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x2 + 1
              app_A(x1,x2) = x1 + x2
              true_0_A() = 1
              filter2_1_A(x1) = x1 + 2
              filter2_2_A(x1,x2) = x1 + x2 + 2
              map_0_A() = 1
              map_1_A(x1) = x1 + 1
              succ_0_A() = 1
              succ_1_A(x1) = x1 + 1
              ack_2_A(x1,x2) = x1 + x2
              |0|_0_A() = 3
              filter_2_A(x1,x2) = x1 + 1
              nil_0_A() = 1
              filter2_4#!_A(x1,x2,x3,x4) = x4 + 2
              false_0!_A() = 0
              filter_2#!_A(x1,x2) = 1
              cons_2!_A(x1,x2) = x1 + x2 + 4
              app!_A(x1,x2) = 3
              true_0!_A() = 0
              filter2_1!_A(x1) = 4
              filter2_2!_A(x1,x2) = 3
              map_0!_A() = 3
              map_1!_A(x1) = 4
              succ_0!_A() = 0
              succ_1!_A(x1) = 7
              ack_2!_A(x1,x2) = x1 + 5
              |0|_0!_A() = 5
              filter_2!_A(x1,x2) = x2 + 1
              nil_0!_A() = 0
        
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app) = []
      pi(true_0) = []
      pi(filter2_1) = [1]
      pi(filter2_2) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(succ_0) = []
      pi(succ_1) = []
      pi(ack_2) = []
      pi(|0|_0) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- Reduction pair.

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

p1: ack_2#(succ_1(x),y) -> ack_2#(x,succ_1(|0|_0()))
p2: ack_2#(succ_1(x),succ_1(y)) -> ack_2#(succ_1(x),y)
p3: ack_2#(succ_1(x),succ_1(y)) -> ack_2#(x,ack_2(succ_1(x),y))

and R consists of:

r1: app(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_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(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r14

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:
          
            ack_2#_A(x1,x2) = 0
            succ_1_A(x1) = 12
            |0|_0_A = 14
            ack_2_A(x1,x2) = max{15, x1}
            ack_2#!_A(x1,x2) = 4
            succ_1!_A(x1) = 14
            |0|_0!_A = 12
            ack_2!_A(x1,x2) = max{5, x1 + 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ack_2#_A(x1,x2) = x1
              succ_1_A(x1) = x1 + 3
              |0|_0_A() = 1
              ack_2_A(x1,x2) = x2 + 4
              ack_2#!_A(x1,x2) = x1
              succ_1!_A(x1) = 1
              |0|_0!_A() = 0
              ack_2!_A(x1,x2) = 2
        
  
    partial status:
  
      pi(ack_2#) = []
      pi(succ_1) = []
      pi(|0|_0) = []
      pi(ack_2) = []

The following rules are removed:

  p1, p3



-- SCC decomposition.

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

p1: ack_2#(succ_1(x),succ_1(y)) -> ack_2#(succ_1(x),y)

and R consists of:

r1: app(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_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(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: ack_2#(succ_1(x),succ_1(y)) -> ack_2#(succ_1(x),y)

and R consists of:

r1: app(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_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(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: 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!):
        
          max/plus interpretation on natural numbers:
          
            ack_2#_A(x1,x2) = x2
            succ_1_A(x1) = max{7, x1}
            ack_2#!_A(x1,x2) = max{0, x1 - 16, x2 - 8}
            succ_1!_A(x1) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ack_2#_A(x1,x2) = x2
              succ_1_A(x1) = x1 + 4
              ack_2#!_A(x1,x2) = x1 + x2
              succ_1!_A(x1) = 9
        
  
    partial status:
  
      pi(ack_2#) = []
      pi(succ_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.