YES
We show the termination of the TRS R:

  app(p(),app(s(),x)) -> x
  app(fact(),|0|()) -> app(s(),|0|())
  app(fact(),app(s(),x)) -> app(app(*(),app(s(),x)),app(fact(),app(p(),app(s(),x))))
  app(app(*(),|0|()),y) -> |0|()
  app(app(*(),app(s(),x)),y) -> app(app(+(),app(app(*(),x),y)),y)
  app(app(+(),x),|0|()) -> x
  app(app(+(),x),app(s(),y)) -> app(s(),app(app(+(),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(*_0(),y) -> *_1(y)
  app(*_1(x1),y) -> *_2(x1,y)
  app(+_0(),y) -> +_1(y)
  app(+_1(x1),y) -> +_2(x1,y)
  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(fact_0(),y) -> fact_1(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(p_0(),y) -> p_1(y)
  app(s_0(),y) -> s_1(y)
  p_1(s_1(x)) -> x
  fact_1(|0|_0()) -> s_1(|0|_0())
  fact_1(s_1(x)) -> *_2(s_1(x),fact_1(p_1(s_1(x))))
  *_2(|0|_0(),y) -> |0|_0()
  *_2(s_1(x),y) -> +_2(*_2(x,y),y)
  +_2(x,|0|_0()) -> x
  +_2(x,s_1(y)) -> s_1(+_2(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#(*_1(x1),y) -> *_2#(x1,y)
p2: app#(+_1(x1),y) -> +_2#(x1,y)
p3: app#(fact_0(),y) -> fact_1#(y)
p4: app#(filter_1(x1),y) -> filter_2#(x1,y)
p5: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p6: app#(map_1(x1),y) -> map_2#(x1,y)
p7: app#(p_0(),y) -> p_1#(y)
p8: fact_1#(s_1(x)) -> *_2#(s_1(x),fact_1(p_1(s_1(x))))
p9: fact_1#(s_1(x)) -> fact_1#(p_1(s_1(x)))
p10: fact_1#(s_1(x)) -> p_1#(s_1(x))
p11: *_2#(s_1(x),y) -> +_2#(*_2(x,y),y)
p12: *_2#(s_1(x),y) -> *_2#(x,y)
p13: +_2#(x,s_1(y)) -> +_2#(x,y)
p14: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p15: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p16: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p17: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p18: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p19: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(fact_0(),y) -> fact_1(y)
r8: app(filter_0(),y) -> filter_1(y)
r9: app(filter_1(x1),y) -> filter_2(x1,y)
r10: app(filter2_0(),y) -> filter2_1(y)
r11: app(filter2_1(x1),y) -> filter2_2(x1,y)
r12: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r13: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r14: app(map_0(),y) -> map_1(y)
r15: app(map_1(x1),y) -> map_2(x1,y)
r16: app(p_0(),y) -> p_1(y)
r17: app(s_0(),y) -> s_1(y)
r18: p_1(s_1(x)) -> x
r19: fact_1(|0|_0()) -> s_1(|0|_0())
r20: fact_1(s_1(x)) -> *_2(s_1(x),fact_1(p_1(s_1(x))))
r21: *_2(|0|_0(),y) -> |0|_0()
r22: *_2(s_1(x),y) -> +_2(*_2(x,y),y)
r23: +_2(x,|0|_0()) -> x
r24: +_2(x,s_1(y)) -> s_1(+_2(x,y))
r25: map_2(f,nil_0()) -> nil_0()
r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r27: filter_2(f,nil_0()) -> nil_0()
r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p4, p5, p6, p14, p15, p16, p17, p18, p19}
  {p9}
  {p12}
  {p13}


-- 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(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(fact_0(),y) -> fact_1(y)
r8: app(filter_0(),y) -> filter_1(y)
r9: app(filter_1(x1),y) -> filter_2(x1,y)
r10: app(filter2_0(),y) -> filter2_1(y)
r11: app(filter2_1(x1),y) -> filter2_2(x1,y)
r12: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r13: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r14: app(map_0(),y) -> map_1(y)
r15: app(map_1(x1),y) -> map_2(x1,y)
r16: app(p_0(),y) -> p_1(y)
r17: app(s_0(),y) -> s_1(y)
r18: p_1(s_1(x)) -> x
r19: fact_1(|0|_0()) -> s_1(|0|_0())
r20: fact_1(s_1(x)) -> *_2(s_1(x),fact_1(p_1(s_1(x))))
r21: *_2(|0|_0(),y) -> |0|_0()
r22: *_2(s_1(x),y) -> +_2(*_2(x,y),y)
r23: +_2(x,|0|_0()) -> x
r24: +_2(x,s_1(y)) -> s_1(+_2(x,y))
r25: map_2(f,nil_0()) -> nil_0()
r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r27: filter_2(f,nil_0()) -> nil_0()
r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r8, r11, r14, r17, r18, r19, r21, r22, r23, r24, r25

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{1, x1 - 15, x2 - 7}
            cons_2_A(x1,x2) = max{x1 + 4, x2 + 6}
            app#_A(x1,x2) = max{1, x1 - 19, x2 - 3}
            map_1_A(x1) = max{18, x1 + 4}
            filter2_3_A(x1,x2,x3) = max{12, x1 - 14, x2 + 4, x3 + 4}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 15, x3 - 15, x4 - 3}
            false_0_A = 33
            filter_2#_A(x1,x2) = max{1, x1 - 15, x2 - 7}
            filter_1_A(x1) = max{12, x1 + 4}
            app_A(x1,x2) = max{35, x1 - 13, x2 + 17}
            true_0_A = 33
            *_0_A = 12
            *_1_A(x1) = x1 + 12
            *_2_A(x1,x2) = max{21, x2 - 1}
            +_0_A = 12
            +_1_A(x1) = x1 + 17
            +_2_A(x1,x2) = max{21, x1, x2 - 20}
            cons_0_A = 12
            cons_1_A(x1) = x1 + 17
            filter_0_A = 12
            filter2_1_A(x1) = max{36, x1 + 17}
            filter2_2_A(x1,x2) = max{24, x1 + 4, x2 + 17}
            map_0_A = 12
            s_0_A = 12
            s_1_A(x1) = max{18, x1 - 22}
            p_1_A(x1) = max{35, x1 + 22}
            fact_1_A(x1) = max{35, x1 - 78}
            |0|_0_A = 2
            map_2_A(x1,x2) = max{x1 + 39, x2 + 17}
            nil_0_A = 40
            map_2#!_A(x1,x2) = max{2, x2 - 1}
            cons_2!_A(x1,x2) = max{53, x1 + 13, x2 + 1}
            app#!_A(x1,x2) = x2 + 7
            map_1!_A(x1) = max{3, x1 - 1}
            filter2_3!_A(x1,x2,x3) = max{53, x1 - 1, x2 - 1, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = 0
            false_0!_A = 55
            filter_2#!_A(x1,x2) = 0
            filter_1!_A(x1) = max{51, x1 + 49}
            app!_A(x1,x2) = max{52, x1 - 25, x2 + 50}
            true_0!_A = 0
            *_0!_A = 1
            *_1!_A(x1) = max{0, x1 - 1}
            *_2!_A(x1,x2) = x2 + 21
            +_0!_A = 1
            +_1!_A(x1) = max{51, x1 - 1}
            +_2!_A(x1,x2) = 0
            cons_0!_A = 53
            cons_1!_A(x1) = max{52, x1 - 1}
            filter_0!_A = 52
            filter2_1!_A(x1) = max{22, x1 + 1}
            filter2_2!_A(x1,x2) = 12
            map_0!_A = 2
            s_0!_A = 0
            s_1!_A(x1) = 0
            p_1!_A(x1) = max{52, x1 - 82}
            fact_1!_A(x1) = max{52, x1 - 41}
            |0|_0!_A = 0
            map_2!_A(x1,x2) = max{49, x2 + 2}
            nil_0!_A = 41
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 2, x2 - 18}
            cons_2_A(x1,x2) = max{17, x1, x2}
            app#_A(x1,x2) = max{1, x1 - 2, x2 - 18}
            map_1_A(x1) = x1 + 1
            filter2_3_A(x1,x2,x3) = max{x1 + 1, x2 + 1, x3 + 1}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 2, x3 - 18, x4 - 18}
            false_0_A = 34
            filter_2#_A(x1,x2) = max{1, x1 - 2, x2 - 18}
            filter_1_A(x1) = x1 + 1
            app_A(x1,x2) = max{69, x1 - 11, x2 + 16}
            true_0_A = 34
            *_0_A = 1
            *_1_A(x1) = 1
            *_2_A(x1,x2) = 20
            +_0_A = 1
            +_1_A(x1) = x1 + 12
            +_2_A(x1,x2) = max{17, x1}
            cons_0_A = 1
            cons_1_A(x1) = x1 + 12
            filter_0_A = 1
            filter2_1_A(x1) = max{10, x1 + 1}
            filter2_2_A(x1,x2) = x2 + 12
            map_0_A = 1
            s_0_A = 1
            s_1_A(x1) = max{17, x1}
            p_1_A(x1) = max{33, x1 + 16}
            fact_1_A(x1) = max{20, x1 - 21}
            |0|_0_A = 20
            map_2_A(x1,x2) = max{69, x1 - 11, x2 + 16}
            nil_0_A = 70
            map_2#!_A(x1,x2) = max{19, x1 + 3, x2 + 2}
            cons_2!_A(x1,x2) = 2
            app#!_A(x1,x2) = max{x1 + 1, x2 - 2}
            map_1!_A(x1) = x1 + 20
            filter2_3!_A(x1,x2,x3) = max{27, x1 - 2, x2 - 1, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = 0
            false_0!_A = 29
            filter_2#!_A(x1,x2) = 0
            filter_1!_A(x1) = max{21, x1 - 29}
            app!_A(x1,x2) = x2 + 25
            true_0!_A = 0
            *_0!_A = 2
            *_1!_A(x1) = max{1, x1 - 53}
            *_2!_A(x1,x2) = max{27, x1 - 52, x2 - 30}
            +_0!_A = 1
            +_1!_A(x1) = max{2, x1 - 1}
            +_2!_A(x1,x2) = x2 + 24
            cons_0!_A = 1
            cons_1!_A(x1) = max{0, x1 - 1}
            filter_0!_A = 1
            filter2_1!_A(x1) = 24
            filter2_2!_A(x1,x2) = max{26, x1 - 1}
            map_0!_A = 21
            s_0!_A = 0
            s_1!_A(x1) = 0
            p_1!_A(x1) = 2
            fact_1!_A(x1) = max{26, x1 - 34}
            |0|_0!_A = 0
            map_2!_A(x1,x2) = max{21, x1 - 1, x2 - 71}
            nil_0!_A = 22
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app#) = [2]
      pi(map_1) = []
      pi(filter2_3) = [3]
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(filter_1) = [1]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(*_0) = []
      pi(*_1) = [1]
      pi(*_2) = [1, 2]
      pi(+_0) = []
      pi(+_1) = [1]
      pi(+_2) = [1, 2]
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(filter_0) = []
      pi(filter2_1) = [1]
      pi(filter2_2) = [1]
      pi(map_0) = []
      pi(s_0) = []
      pi(s_1) = [1]
      pi(p_1) = [1]
      pi(fact_1) = [1]
      pi(|0|_0) = []
      pi(map_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3, p4, p6, p7, p8



-- SCC decomposition.

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

p1: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p2: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(fact_0(),y) -> fact_1(y)
r8: app(filter_0(),y) -> filter_1(y)
r9: app(filter_1(x1),y) -> filter_2(x1,y)
r10: app(filter2_0(),y) -> filter2_1(y)
r11: app(filter2_1(x1),y) -> filter2_2(x1,y)
r12: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r13: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r14: app(map_0(),y) -> map_1(y)
r15: app(map_1(x1),y) -> map_2(x1,y)
r16: app(p_0(),y) -> p_1(y)
r17: app(s_0(),y) -> s_1(y)
r18: p_1(s_1(x)) -> x
r19: fact_1(|0|_0()) -> s_1(|0|_0())
r20: fact_1(s_1(x)) -> *_2(s_1(x),fact_1(p_1(s_1(x))))
r21: *_2(|0|_0(),y) -> |0|_0()
r22: *_2(s_1(x),y) -> +_2(*_2(x,y),y)
r23: +_2(x,|0|_0()) -> x
r24: +_2(x,s_1(y)) -> s_1(+_2(x,y))
r25: map_2(f,nil_0()) -> nil_0()
r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r27: filter_2(f,nil_0()) -> nil_0()
r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r30: 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: fact_1#(s_1(x)) -> fact_1#(p_1(s_1(x)))

and R consists of:

r1: app(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(fact_0(),y) -> fact_1(y)
r8: app(filter_0(),y) -> filter_1(y)
r9: app(filter_1(x1),y) -> filter_2(x1,y)
r10: app(filter2_0(),y) -> filter2_1(y)
r11: app(filter2_1(x1),y) -> filter2_2(x1,y)
r12: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r13: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r14: app(map_0(),y) -> map_1(y)
r15: app(map_1(x1),y) -> map_2(x1,y)
r16: app(p_0(),y) -> p_1(y)
r17: app(s_0(),y) -> s_1(y)
r18: p_1(s_1(x)) -> x
r19: fact_1(|0|_0()) -> s_1(|0|_0())
r20: fact_1(s_1(x)) -> *_2(s_1(x),fact_1(p_1(s_1(x))))
r21: *_2(|0|_0(),y) -> |0|_0()
r22: *_2(s_1(x),y) -> +_2(*_2(x,y),y)
r23: +_2(x,|0|_0()) -> x
r24: +_2(x,s_1(y)) -> s_1(+_2(x,y))
r25: map_2(f,nil_0()) -> nil_0()
r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r27: filter_2(f,nil_0()) -> nil_0()
r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r18

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:
          
            fact_1#_A(x1) = max{0, x1 - 45}
            s_1_A(x1) = max{44, x1 + 32}
            p_1_A(x1) = max{33, x1 - 4}
            fact_1#!_A(x1) = max{9, x1 - 34}
            s_1!_A(x1) = max{7, x1 - 86}
            p_1!_A(x1) = max{8, x1 - 127}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            fact_1#_A(x1) = max{0, x1 - 38}
            s_1_A(x1) = max{5, x1 + 1}
            p_1_A(x1) = max{37, x1 - 1}
            fact_1#!_A(x1) = max{11, x1 - 39}
            s_1!_A(x1) = max{8, x1 - 39}
            p_1!_A(x1) = max{10, x1 - 45}
        
  
    partial status:
  
      pi(fact_1#) = [1]
      pi(s_1) = [1]
      pi(p_1) = [1]

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: *_2#(s_1(x),y) -> *_2#(x,y)

and R consists of:

r1: app(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(fact_0(),y) -> fact_1(y)
r8: app(filter_0(),y) -> filter_1(y)
r9: app(filter_1(x1),y) -> filter_2(x1,y)
r10: app(filter2_0(),y) -> filter2_1(y)
r11: app(filter2_1(x1),y) -> filter2_2(x1,y)
r12: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r13: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r14: app(map_0(),y) -> map_1(y)
r15: app(map_1(x1),y) -> map_2(x1,y)
r16: app(p_0(),y) -> p_1(y)
r17: app(s_0(),y) -> s_1(y)
r18: p_1(s_1(x)) -> x
r19: fact_1(|0|_0()) -> s_1(|0|_0())
r20: fact_1(s_1(x)) -> *_2(s_1(x),fact_1(p_1(s_1(x))))
r21: *_2(|0|_0(),y) -> |0|_0()
r22: *_2(s_1(x),y) -> +_2(*_2(x,y),y)
r23: +_2(x,|0|_0()) -> x
r24: +_2(x,s_1(y)) -> s_1(+_2(x,y))
r25: map_2(f,nil_0()) -> nil_0()
r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r27: filter_2(f,nil_0()) -> nil_0()
r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r30: 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:
          
            *_2#_A(x1,x2) = 0
            s_1_A(x1) = max{5, x1 - 5}
            *_2#!_A(x1,x2) = max{1, x1 - 22}
            s_1!_A(x1) = max{2, x1 - 21}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            *_2#_A(x1,x2) = 0
            s_1_A(x1) = max{12, x1 - 12}
            *_2#!_A(x1,x2) = max{11, x1 - 13, x2 - 1}
            s_1!_A(x1) = max{10, x1 - 101}
        
  
    partial status:
  
      pi(*_2#) = [1]
      pi(s_1) = [1]

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: +_2#(x,s_1(y)) -> +_2#(x,y)

and R consists of:

r1: app(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(fact_0(),y) -> fact_1(y)
r8: app(filter_0(),y) -> filter_1(y)
r9: app(filter_1(x1),y) -> filter_2(x1,y)
r10: app(filter2_0(),y) -> filter2_1(y)
r11: app(filter2_1(x1),y) -> filter2_2(x1,y)
r12: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r13: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r14: app(map_0(),y) -> map_1(y)
r15: app(map_1(x1),y) -> map_2(x1,y)
r16: app(p_0(),y) -> p_1(y)
r17: app(s_0(),y) -> s_1(y)
r18: p_1(s_1(x)) -> x
r19: fact_1(|0|_0()) -> s_1(|0|_0())
r20: fact_1(s_1(x)) -> *_2(s_1(x),fact_1(p_1(s_1(x))))
r21: *_2(|0|_0(),y) -> |0|_0()
r22: *_2(s_1(x),y) -> +_2(*_2(x,y),y)
r23: +_2(x,|0|_0()) -> x
r24: +_2(x,s_1(y)) -> s_1(+_2(x,y))
r25: map_2(f,nil_0()) -> nil_0()
r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r27: filter_2(f,nil_0()) -> nil_0()
r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r30: 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:
          
            +_2#_A(x1,x2) = 0
            s_1_A(x1) = max{8, x1 - 8}
            +_2#!_A(x1,x2) = max{1, x2 - 95}
            s_1!_A(x1) = max{4, x1 - 94}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            +_2#_A(x1,x2) = 0
            s_1_A(x1) = max{12, x1 - 12}
            +_2#!_A(x1,x2) = max{5, x1 - 1, x2 - 13}
            s_1!_A(x1) = max{4, x1 - 72}
        
  
    partial status:
  
      pi(+_2#) = [2]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.