YES
We show the termination of the TRS R:

  app(app(times(),x),|0|()) -> |0|()
  app(app(times(),x),app(s(),y)) -> app(app(plus(),app(app(times(),x),y)),x)
  app(app(plus(),x),|0|()) -> x
  app(app(plus(),|0|()),x) -> x
  app(app(plus(),x),app(s(),y)) -> app(s(),app(app(plus(),x),y))
  app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),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(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(plus_0(),y) -> plus_1(y)
  app(plus_1(x1),y) -> plus_2(x1,y)
  app(s_0(),y) -> s_1(y)
  app(times_0(),y) -> times_1(y)
  app(times_1(x1),y) -> times_2(x1,y)
  times_2(x,|0|_0()) -> |0|_0()
  times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
  plus_2(x,|0|_0()) -> x
  plus_2(|0|_0(),x) -> x
  plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
  plus_2(s_1(x),y) -> s_1(plus_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#(filter_1(x1),y) -> filter_2#(x1,y)
p2: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: app#(plus_1(x1),y) -> plus_2#(x1,y)
p5: app#(times_1(x1),y) -> times_2#(x1,y)
p6: times_2#(x,s_1(y)) -> plus_2#(times_2(x,y),x)
p7: times_2#(x,s_1(y)) -> times_2#(x,y)
p8: plus_2#(x,s_1(y)) -> plus_2#(x,y)
p9: plus_2#(s_1(x),y) -> plus_2#(x,y)
p10: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p11: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p12: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p13: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p14: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p15: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p10, p11, p12, p13, p14, p15}
  {p7}
  {p8, p9}


-- Reduction pair.

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

p1: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p7: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
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(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r19, r24

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{1, x1 - 17, x2 - 18}
            filter_1_A(x1) = max{7, x1 + 3}
            filter_2#_A(x1,x2) = max{1, x1 - 17, x2 - 23}
            cons_2_A(x1,x2) = max{22, x1 + 5, x2}
            map_1_A(x1) = max{5, x1 + 3}
            map_2#_A(x1,x2) = max{1, x1 - 14, x2 - 23}
            filter2_3_A(x1,x2,x3) = max{x1 + 3, x2 + 3, x3 + 16}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 17, x3 - 18, x4 - 23}
            false_0_A = 44
            app_A(x1,x2) = max{44, x1 - 12, x2 + 15}
            true_0_A = 20
            plus_2_A(x1,x2) = max{98, x1 + 2, x2 + 3}
            |0|_0_A = 46
            filter_2_A(x1,x2) = x2
            nil_0_A = 40
            app#!_A(x1,x2) = max{0, x2 - 4}
            filter_1!_A(x1) = max{0, x1 - 1}
            filter_2#!_A(x1,x2) = max{0, x2 - 4}
            cons_2!_A(x1,x2) = max{x1 + 2, x2 + 5}
            map_1!_A(x1) = max{1, x1 - 2}
            map_2#!_A(x1,x2) = max{0, x2 - 6}
            filter2_3!_A(x1,x2,x3) = 0
            filter2_4#!_A(x1,x2,x3,x4) = max{0, x4 - 4}
            false_0!_A = 75
            app!_A(x1,x2) = max{x1 - 6, x2 + 73}
            true_0!_A = 6
            plus_2!_A(x1,x2) = 1
            |0|_0!_A = 3
            filter_2!_A(x1,x2) = max{74, x2 + 5}
            nil_0!_A = 46
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{1, x2 - 4}
            filter_1_A(x1) = max{24, x1 - 6}
            filter_2#_A(x1,x2) = max{1, x2 - 12}
            cons_2_A(x1,x2) = max{11, x1 + 8, x2}
            map_1_A(x1) = max{24, x1 - 6}
            map_2#_A(x1,x2) = max{1, x2 - 12}
            filter2_3_A(x1,x2,x3) = max{24, x1 - 6, x2 + 1, x3 - 6}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x4 - 12}
            false_0_A = 7
            app_A(x1,x2) = max{23, x1 - 35, x2 + 3}
            true_0_A = 8
            plus_2_A(x1,x2) = max{29, x2 + 4}
            |0|_0_A = 25
            filter_2_A(x1,x2) = max{24, x1 - 34, x2 + 13}
            nil_0_A = 25
            app#!_A(x1,x2) = max{38, x2 + 36}
            filter_1!_A(x1) = max{3, x1 - 5}
            filter_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = 0
            map_1!_A(x1) = x1 + 1
            map_2#!_A(x1,x2) = 37
            filter2_3!_A(x1,x2,x3) = max{x1 - 1, x2 + 37, x3 - 5}
            filter2_4#!_A(x1,x2,x3,x4) = 0
            false_0!_A = 1
            app!_A(x1,x2) = max{28, x1 + 2, x2 - 4}
            true_0!_A = 1
            plus_2!_A(x1,x2) = 27
            |0|_0!_A = 1
            filter_2!_A(x1,x2) = x1
            nil_0!_A = 0
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = []
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(map_1) = []
      pi(map_2#) = []
      pi(filter2_3) = [1]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(plus_2) = [1, 2]
      pi(|0|_0) = []
      pi(filter_2) = [1]
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3, p5, p6



-- 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(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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{4, x1 - 32, x2 - 33}
            map_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = max{x1 + 2, x2 - 42}
        
        2. 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{5, x1 - 7, x2 - 2}
            map_2#!_A(x1,x2) = max{1, x1 - 1, x2 - 49}
            cons_2!_A(x1,x2) = max{4, x1 - 68, x2 - 48}
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [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(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r2, r5, r12, r15, r16, r17, r18, r19, r20, r21, r24

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}
            false_0_A = 3
            filter_2#_A(x1,x2) = max{1, x1 - 1}
            cons_2_A(x1,x2) = max{16, x1 - 11, x2 - 12}
            app_A(x1,x2) = max{11, x1 - 5, x2 + 3}
            true_0_A = 3
            cons_1_A(x1) = max{21, x1 + 3}
            filter2_0_A = 0
            filter2_1_A(x1) = max{3, x1 - 6}
            plus_1_A(x1) = x1 + 18
            plus_2_A(x1,x2) = max{13, x1, x2 + 3}
            times_1_A(x1) = x1 + 34
            times_2_A(x1,x2) = x1 + 14
            |0|_0_A = 14
            s_1_A(x1) = 14
            filter_2_A(x1,x2) = max{31, x1 + 15, x2 - 1}
            nil_0_A = 16
            filter2_4#!_A(x1,x2,x3,x4) = 6
            false_0!_A = 3
            filter_2#!_A(x1,x2) = 4
            cons_2!_A(x1,x2) = max{x1 + 7, x2 - 14}
            app!_A(x1,x2) = max{x1 + 5, x2 + 36}
            true_0!_A = 1
            cons_1!_A(x1) = max{37, x1 - 1}
            filter2_0!_A = 5
            filter2_1!_A(x1) = max{6, x1 - 3}
            plus_1!_A(x1) = max{0, x1 - 1}
            plus_2!_A(x1,x2) = 0
            times_1!_A(x1) = 37
            times_2!_A(x1,x2) = max{38, x1 + 2}
            |0|_0!_A = 3
            s_1!_A(x1) = 0
            filter_2!_A(x1,x2) = max{x1 + 3, x2 + 17}
            nil_0!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            filter2_4#_A(x1,x2,x3,x4) = max{32, x2 - 1, x3 - 1, x4 + 15}
            false_0_A = 14
            filter_2#_A(x1,x2) = max{32, x1 - 1, x2 + 15}
            cons_2_A(x1,x2) = max{16, x1 - 15, x2}
            app_A(x1,x2) = max{x1 - 13, x2 + 14}
            true_0_A = 14
            cons_1_A(x1) = max{29, x1 + 12}
            filter2_0_A = 0
            filter2_1_A(x1) = x1
            plus_1_A(x1) = x1 + 14
            plus_2_A(x1,x2) = max{6, x1, x2 + 1}
            times_1_A(x1) = x1 + 14
            times_2_A(x1,x2) = max{6, x1 + 1}
            |0|_0_A = 2
            s_1_A(x1) = max{4, x1}
            filter_2_A(x1,x2) = max{16, x1 - 34, x2 + 14}
            nil_0_A = 17
            filter2_4#!_A(x1,x2,x3,x4) = max{3, x3 - 31}
            false_0!_A = 1
            filter_2#!_A(x1,x2) = 2
            cons_2!_A(x1,x2) = x1 + 4
            app!_A(x1,x2) = max{x1 - 1, x2 + 15}
            true_0!_A = 1
            cons_1!_A(x1) = max{16, x1 - 1}
            filter2_0!_A = 17
            filter2_1!_A(x1) = max{16, x1 - 17}
            plus_1!_A(x1) = max{16, x1 - 8}
            plus_2!_A(x1,x2) = max{14, x1 - 7, x2 + 1}
            times_1!_A(x1) = 16
            times_2!_A(x1,x2) = 0
            |0|_0!_A = 0
            s_1!_A(x1) = max{0, x1 - 14}
            filter_2!_A(x1,x2) = 0
            nil_0!_A = 1
        
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(cons_1) = [1]
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(plus_1) = [1]
      pi(plus_2) = [1, 2]
      pi(times_1) = [1]
      pi(times_2) = [1, 2]
      pi(|0|_0) = []
      pi(s_1) = [1]
      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: times_2#(x,s_1(y)) -> times_2#(x,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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:
          
            times_2#_A(x1,x2) = 0
            s_1_A(x1) = max{4, x1 - 4}
            times_2#!_A(x1,x2) = max{0, x2 - 16}
            s_1!_A(x1) = max{4, x1 - 15}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times_2#_A(x1,x2) = 0
            s_1_A(x1) = max{4, x1 - 4}
            times_2#!_A(x1,x2) = max{7, x1 - 1, x2 - 7}
            s_1!_A(x1) = max{6, x1 - 6}
        
  
    partial status:
  
      pi(times_2#) = [2]
      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: plus_2#(x,s_1(y)) -> plus_2#(x,y)
p2: plus_2#(s_1(x),y) -> plus_2#(x,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{7, x1 - 7}
            plus_2#!_A(x1,x2) = max{0, x1 - 8}
            s_1!_A(x1) = max{0, x1 - 7}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{4, x1 - 4}
            plus_2#!_A(x1,x2) = max{0, x1 - 20}
            s_1!_A(x1) = max{4, x1 - 20}
        
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: plus_2#(x,s_1(y)) -> plus_2#(x,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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: plus_2#(x,s_1(y)) -> plus_2#(x,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{4, x1 - 4}
            plus_2#!_A(x1,x2) = max{1, x2 - 38}
            s_1!_A(x1) = max{4, x1 - 37}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{7, x1 - 7}
            plus_2#!_A(x1,x2) = max{5, x1 - 1, x2 - 8}
            s_1!_A(x1) = max{4, x1 - 54}
        
  
    partial status:
  
      pi(plus_2#) = [2]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.