YES
We show the termination of the TRS R:

  |app'|(|app'|(minus(),x),|0|()) -> x
  |app'|(|app'|(minus(),|app'|(s(),x)),|app'|(s(),y)) -> |app'|(|app'|(minus(),x),y)
  |app'|(|app'|(minus(),|app'|(|app'|(minus(),x),y)),z) -> |app'|(|app'|(minus(),x),|app'|(|app'|(plus(),y),z))
  |app'|(|app'|(quot(),|0|()),|app'|(s(),y)) -> |0|()
  |app'|(|app'|(quot(),|app'|(s(),x)),|app'|(s(),y)) -> |app'|(s(),|app'|(|app'|(quot(),|app'|(|app'|(minus(),x),y)),|app'|(s(),y)))
  |app'|(|app'|(plus(),|0|()),y) -> y
  |app'|(|app'|(plus(),|app'|(s(),x)),y) -> |app'|(s(),|app'|(|app'|(plus(),x),y))
  |app'|(|app'|(app(),nil()),k) -> k
  |app'|(|app'|(app(),l),nil()) -> l
  |app'|(|app'|(app(),|app'|(|app'|(cons(),x),l)),k) -> |app'|(|app'|(cons(),x),|app'|(|app'|(app(),l),k))
  |app'|(sum(),|app'|(|app'|(cons(),x),nil())) -> |app'|(|app'|(cons(),x),nil())
  |app'|(sum(),|app'|(|app'|(cons(),x),|app'|(|app'|(cons(),y),l))) -> |app'|(sum(),|app'|(|app'|(cons(),|app'|(|app'|(plus(),x),y)),l))
  |app'|(sum(),|app'|(|app'|(app(),l),|app'|(|app'|(cons(),x),|app'|(|app'|(cons(),y),k)))) -> |app'|(sum(),|app'|(|app'|(app(),l),|app'|(sum(),|app'|(|app'|(cons(),x),|app'|(|app'|(cons(),y),k)))))
  |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'|(app_0(),y) -> app_1(y)
  |app'|(app_1(x1),y) -> app_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'|(minus_0(),y) -> minus_1(y)
  |app'|(minus_1(x1),y) -> minus_2(x1,y)
  |app'|(plus_0(),y) -> plus_1(y)
  |app'|(plus_1(x1),y) -> plus_2(x1,y)
  |app'|(quot_0(),y) -> quot_1(y)
  |app'|(quot_1(x1),y) -> quot_2(x1,y)
  |app'|(s_0(),y) -> s_1(y)
  |app'|(sum_0(),y) -> sum_1(y)
  minus_2(x,|0|_0()) -> x
  minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
  minus_2(minus_2(x,y),z) -> minus_2(x,plus_2(y,z))
  quot_2(|0|_0(),s_1(y)) -> |0|_0()
  quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
  plus_2(|0|_0(),y) -> y
  plus_2(s_1(x),y) -> s_1(plus_2(x,y))
  app_2(nil_0(),k) -> k
  app_2(l,nil_0()) -> l
  app_2(cons_2(x,l),k) -> cons_2(x,app_2(l,k))
  sum_1(cons_2(x,nil_0())) -> cons_2(x,nil_0())
  sum_1(cons_2(x,cons_2(y,l))) -> sum_1(cons_2(plus_2(x,y),l))
  sum_1(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
  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'|#(app_1(x1),y) -> app_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'|#(minus_1(x1),y) -> minus_2#(x1,y)
p6: |app'|#(plus_1(x1),y) -> plus_2#(x1,y)
p7: |app'|#(quot_1(x1),y) -> quot_2#(x1,y)
p8: |app'|#(sum_0(),y) -> sum_1#(y)
p9: minus_2#(s_1(x),s_1(y)) -> minus_2#(x,y)
p10: minus_2#(minus_2(x,y),z) -> minus_2#(x,plus_2(y,z))
p11: minus_2#(minus_2(x,y),z) -> plus_2#(y,z)
p12: quot_2#(s_1(x),s_1(y)) -> quot_2#(minus_2(x,y),s_1(y))
p13: quot_2#(s_1(x),s_1(y)) -> minus_2#(x,y)
p14: plus_2#(s_1(x),y) -> plus_2#(x,y)
p15: app_2#(cons_2(x,l),k) -> app_2#(l,k)
p16: sum_1#(cons_2(x,cons_2(y,l))) -> sum_1#(cons_2(plus_2(x,y),l))
p17: sum_1#(cons_2(x,cons_2(y,l))) -> plus_2#(x,y)
p18: sum_1#(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1#(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
p19: sum_1#(app_2(l,cons_2(x,cons_2(y,k)))) -> app_2#(l,sum_1(cons_2(x,cons_2(y,k))))
p20: sum_1#(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1#(cons_2(x,cons_2(y,k)))
p21: map_2#(f,cons_2(x,xs)) -> |app'|#(f,x)
p22: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p23: filter_2#(f,cons_2(x,xs)) -> filter2_4#(|app'|(f,x),f,x,xs)
p24: filter_2#(f,cons_2(x,xs)) -> |app'|#(f,x)
p25: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p26: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: |app'|(app_0(),y) -> app_1(y)
r2: |app'|(app_1(x1),y) -> app_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'|(minus_0(),y) -> minus_1(y)
r14: |app'|(minus_1(x1),y) -> minus_2(x1,y)
r15: |app'|(plus_0(),y) -> plus_1(y)
r16: |app'|(plus_1(x1),y) -> plus_2(x1,y)
r17: |app'|(quot_0(),y) -> quot_1(y)
r18: |app'|(quot_1(x1),y) -> quot_2(x1,y)
r19: |app'|(s_0(),y) -> s_1(y)
r20: |app'|(sum_0(),y) -> sum_1(y)
r21: minus_2(x,|0|_0()) -> x
r22: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r23: minus_2(minus_2(x,y),z) -> minus_2(x,plus_2(y,z))
r24: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r25: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r26: plus_2(|0|_0(),y) -> y
r27: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r28: app_2(nil_0(),k) -> k
r29: app_2(l,nil_0()) -> l
r30: app_2(cons_2(x,l),k) -> cons_2(x,app_2(l,k))
r31: sum_1(cons_2(x,nil_0())) -> cons_2(x,nil_0())
r32: sum_1(cons_2(x,cons_2(y,l))) -> sum_1(cons_2(plus_2(x,y),l))
r33: sum_1(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
r34: map_2(f,nil_0()) -> nil_0()
r35: map_2(f,cons_2(x,xs)) -> cons_2(|app'|(f,x),map_2(f,xs))
r36: filter_2(f,nil_0()) -> nil_0()
r37: filter_2(f,cons_2(x,xs)) -> filter2_4(|app'|(f,x),f,x,xs)
r38: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r39: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p21, p22, p23, p24, p25, p26}
  {p18}
  {p15}
  {p12}
  {p9, p10}
  {p16}
  {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'|(app_0(),y) -> app_1(y)
r2: |app'|(app_1(x1),y) -> app_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'|(minus_0(),y) -> minus_1(y)
r14: |app'|(minus_1(x1),y) -> minus_2(x1,y)
r15: |app'|(plus_0(),y) -> plus_1(y)
r16: |app'|(plus_1(x1),y) -> plus_2(x1,y)
r17: |app'|(quot_0(),y) -> quot_1(y)
r18: |app'|(quot_1(x1),y) -> quot_2(x1,y)
r19: |app'|(s_0(),y) -> s_1(y)
r20: |app'|(sum_0(),y) -> sum_1(y)
r21: minus_2(x,|0|_0()) -> x
r22: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r23: minus_2(minus_2(x,y),z) -> minus_2(x,plus_2(y,z))
r24: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r25: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r26: plus_2(|0|_0(),y) -> y
r27: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r28: app_2(nil_0(),k) -> k
r29: app_2(l,nil_0()) -> l
r30: app_2(cons_2(x,l),k) -> cons_2(x,app_2(l,k))
r31: sum_1(cons_2(x,nil_0())) -> cons_2(x,nil_0())
r32: sum_1(cons_2(x,cons_2(y,l))) -> sum_1(cons_2(plus_2(x,y),l))
r33: sum_1(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
r34: map_2(f,nil_0()) -> nil_0()
r35: map_2(f,cons_2(x,xs)) -> cons_2(|app'|(f,x),map_2(f,xs))
r36: filter_2(f,nil_0()) -> nil_0()
r37: filter_2(f,cons_2(x,xs)) -> filter2_4(|app'|(f,x),f,x,xs)
r38: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r39: 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) = max{1, x1 - 3, x2 - 71}
            cons_2_A(x1,x2) = max{70, x1 + 19, x2 + 19}
            |app'|#_A(x1,x2) = max{1, x1 - 4, x2 - 52}
            map_1_A(x1) = max{3, x1 + 1}
            filter2_3_A(x1,x2,x3) = max{x1 - 2, x2 + 1, x3 + 3}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 4, x3 - 51, x4 - 52}
            false_0_A = 2
            filter_2#_A(x1,x2) = max{1, x1 - 4, x2 - 70}
            filter_1_A(x1) = max{3, x1 + 1}
            |app'|_A(x1,x2) = max{2, x1 - 291, x2 - 291}
            true_0_A = 0
            map_2#!_A(x1,x2) = max{9, x1 + 1}
            cons_2!_A(x1,x2) = max{72, x1 + 8, x2 + 2}
            |app'|#!_A(x1,x2) = 0
            map_1!_A(x1) = max{10, x1 + 2}
            filter2_3!_A(x1,x2,x3) = max{0, x1 - 1, x2 - 1, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = 0
            false_0!_A = 302
            filter_2#!_A(x1,x2) = 0
            filter_1!_A(x1) = max{2, x1 - 1}
            |app'|!_A(x1,x2) = 0
            true_0!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 121, x2 - 79}
            cons_2_A(x1,x2) = max{78, x1 + 7, x2 + 39}
            |app'|#_A(x1,x2) = max{1, x1 - 122, x2 - 72}
            map_1_A(x1) = max{119, x1 + 1}
            filter2_3_A(x1,x2,x3) = max{121, x1 - 1, x2 + 1, x3 - 1}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 121, x3 - 123, x4 - 72}
            false_0_A = 2
            filter_2#_A(x1,x2) = max{1, x1 - 121, x2 - 78}
            filter_1_A(x1) = max{3, x1 + 1}
            |app'|_A(x1,x2) = max{2, x1 - 123, x2 - 48}
            true_0_A = 0
            map_2#!_A(x1,x2) = max{124, x1 - 120}
            cons_2!_A(x1,x2) = max{123, x1 + 6, x2 + 117}
            |app'|#!_A(x1,x2) = x2 + 5
            map_1!_A(x1) = max{125, x1 - 119}
            filter2_3!_A(x1,x2,x3) = max{13, x1 - 1, x2 - 120, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = x4 + 4
            false_0!_A = 1
            filter_2#!_A(x1,x2) = max{3, x2 + 1}
            filter_1!_A(x1) = max{0, x1 - 1}
            |app'|!_A(x1,x2) = max{11, x1 - 120, x2 + 9}
            true_0!_A = 4
        
  
    partial status:
  
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(|app'|#) = [1]
      pi(map_1) = [1]
      pi(filter2_3) = [2, 3]
      pi(filter2_4#) = [2]
      pi(false_0) = []
      pi(filter_2#) = [1]
      pi(filter_1) = [1]
      pi(|app'|) = [2]
      pi(true_0) = []

The following rules are removed:

  p1, p2, p4, p5, p6, p7, p8, p9



-- 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)

and R consists of:

r1: |app'|(app_0(),y) -> app_1(y)
r2: |app'|(app_1(x1),y) -> app_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'|(minus_0(),y) -> minus_1(y)
r14: |app'|(minus_1(x1),y) -> minus_2(x1,y)
r15: |app'|(plus_0(),y) -> plus_1(y)
r16: |app'|(plus_1(x1),y) -> plus_2(x1,y)
r17: |app'|(quot_0(),y) -> quot_1(y)
r18: |app'|(quot_1(x1),y) -> quot_2(x1,y)
r19: |app'|(s_0(),y) -> s_1(y)
r20: |app'|(sum_0(),y) -> sum_1(y)
r21: minus_2(x,|0|_0()) -> x
r22: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r23: minus_2(minus_2(x,y),z) -> minus_2(x,plus_2(y,z))
r24: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r25: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r26: plus_2(|0|_0(),y) -> y
r27: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r28: app_2(nil_0(),k) -> k
r29: app_2(l,nil_0()) -> l
r30: app_2(cons_2(x,l),k) -> cons_2(x,app_2(l,k))
r31: sum_1(cons_2(x,nil_0())) -> cons_2(x,nil_0())
r32: sum_1(cons_2(x,cons_2(y,l))) -> sum_1(cons_2(plus_2(x,y),l))
r33: sum_1(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
r34: map_2(f,nil_0()) -> nil_0()
r35: map_2(f,cons_2(x,xs)) -> cons_2(|app'|(f,x),map_2(f,xs))
r36: filter_2(f,nil_0()) -> nil_0()
r37: filter_2(f,cons_2(x,xs)) -> filter2_4(|app'|(f,x),f,x,xs)
r38: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r39: 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: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: |app'|(app_0(),y) -> app_1(y)
r2: |app'|(app_1(x1),y) -> app_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'|(minus_0(),y) -> minus_1(y)
r14: |app'|(minus_1(x1),y) -> minus_2(x1,y)
r15: |app'|(plus_0(),y) -> plus_1(y)
r16: |app'|(plus_1(x1),y) -> plus_2(x1,y)
r17: |app'|(quot_0(),y) -> quot_1(y)
r18: |app'|(quot_1(x1),y) -> quot_2(x1,y)
r19: |app'|(s_0(),y) -> s_1(y)
r20: |app'|(sum_0(),y) -> sum_1(y)
r21: minus_2(x,|0|_0()) -> x
r22: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r23: minus_2(minus_2(x,y),z) -> minus_2(x,plus_2(y,z))
r24: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r25: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r26: plus_2(|0|_0(),y) -> y
r27: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r28: app_2(nil_0(),k) -> k
r29: app_2(l,nil_0()) -> l
r30: app_2(cons_2(x,l),k) -> cons_2(x,app_2(l,k))
r31: sum_1(cons_2(x,nil_0())) -> cons_2(x,nil_0())
r32: sum_1(cons_2(x,cons_2(y,l))) -> sum_1(cons_2(plus_2(x,y),l))
r33: sum_1(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
r34: map_2(f,nil_0()) -> nil_0()
r35: map_2(f,cons_2(x,xs)) -> cons_2(|app'|(f,x),map_2(f,xs))
r36: filter_2(f,nil_0()) -> nil_0()
r37: filter_2(f,cons_2(x,xs)) -> filter2_4(|app'|(f,x),f,x,xs)
r38: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r39: 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) = max{0, x2 - 9}
            cons_2_A(x1,x2) = max{x1 - 98, x2 + 8}
            map_2#!_A(x1,x2) = max{8, x1 - 1, x2 - 9}
            cons_2!_A(x1,x2) = max{7, x1 - 327, x2 - 238}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{0, x2 - 2}
            cons_2_A(x1,x2) = max{49, x1 - 17, x2 + 1}
            map_2#!_A(x1,x2) = max{1, x1 - 1, x2 - 50}
            cons_2!_A(x1,x2) = max{0, x2 - 51}
        
  
    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: sum_1#(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1#(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))

and R consists of:

r1: |app'|(app_0(),y) -> app_1(y)
r2: |app'|(app_1(x1),y) -> app_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'|(minus_0(),y) -> minus_1(y)
r14: |app'|(minus_1(x1),y) -> minus_2(x1,y)
r15: |app'|(plus_0(),y) -> plus_1(y)
r16: |app'|(plus_1(x1),y) -> plus_2(x1,y)
r17: |app'|(quot_0(),y) -> quot_1(y)
r18: |app'|(quot_1(x1),y) -> quot_2(x1,y)
r19: |app'|(s_0(),y) -> s_1(y)
r20: |app'|(sum_0(),y) -> sum_1(y)
r21: minus_2(x,|0|_0()) -> x
r22: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r23: minus_2(minus_2(x,y),z) -> minus_2(x,plus_2(y,z))
r24: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r25: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r26: plus_2(|0|_0(),y) -> y
r27: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r28: app_2(nil_0(),k) -> k
r29: app_2(l,nil_0()) -> l
r30: app_2(cons_2(x,l),k) -> cons_2(x,app_2(l,k))
r31: sum_1(cons_2(x,nil_0())) -> cons_2(x,nil_0())
r32: sum_1(cons_2(x,cons_2(y,l))) -> sum_1(cons_2(plus_2(x,y),l))
r33: sum_1(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
r34: map_2(f,nil_0()) -> nil_0()
r35: map_2(f,cons_2(x,xs)) -> cons_2(|app'|(f,x),map_2(f,xs))
r36: filter_2(f,nil_0()) -> nil_0()
r37: filter_2(f,cons_2(x,xs)) -> filter2_4(|app'|(f,x),f,x,xs)
r38: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r39: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r28, r29, r30, r31, r32, r33

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:
          
            sum_1#_A(x1) = 0
            app_2_A(x1,x2) = max{142, x1 + 61, x2 + 18}
            cons_2_A(x1,x2) = 77
            sum_1_A(x1) = x1 + 47
            nil_0_A = 79
            plus_2_A(x1,x2) = x1 + 68
            sum_1#!_A(x1) = max{17, x1 - 44}
            app_2!_A(x1,x2) = max{x1 + 66, x2 - 58}
            cons_2!_A(x1,x2) = 16
            sum_1!_A(x1) = max{2, x1 - 61}
            nil_0!_A = 3
            plus_2!_A(x1,x2) = max{0, x1 - 30}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum_1#_A(x1) = max{0, x1 - 22}
            app_2_A(x1,x2) = max{50, x1 + 23, x2 + 2}
            cons_2_A(x1,x2) = 22
            sum_1_A(x1) = 47
            nil_0_A = 9
            plus_2_A(x1,x2) = max{18, x1 - 36}
            sum_1#!_A(x1) = 0
            app_2!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = max{24, x2 + 16}
            sum_1!_A(x1) = 37
            nil_0!_A = 26
            plus_2!_A(x1,x2) = max{24, x2}
        
  
    partial status:
  
      pi(sum_1#) = [1]
      pi(app_2) = [1, 2]
      pi(cons_2) = []
      pi(sum_1) = []
      pi(nil_0) = []
      pi(plus_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: app_2#(cons_2(x,l),k) -> app_2#(l,k)

and R consists of:

r1: |app'|(app_0(),y) -> app_1(y)
r2: |app'|(app_1(x1),y) -> app_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'|(minus_0(),y) -> minus_1(y)
r14: |app'|(minus_1(x1),y) -> minus_2(x1,y)
r15: |app'|(plus_0(),y) -> plus_1(y)
r16: |app'|(plus_1(x1),y) -> plus_2(x1,y)
r17: |app'|(quot_0(),y) -> quot_1(y)
r18: |app'|(quot_1(x1),y) -> quot_2(x1,y)
r19: |app'|(s_0(),y) -> s_1(y)
r20: |app'|(sum_0(),y) -> sum_1(y)
r21: minus_2(x,|0|_0()) -> x
r22: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r23: minus_2(minus_2(x,y),z) -> minus_2(x,plus_2(y,z))
r24: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r25: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r26: plus_2(|0|_0(),y) -> y
r27: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r28: app_2(nil_0(),k) -> k
r29: app_2(l,nil_0()) -> l
r30: app_2(cons_2(x,l),k) -> cons_2(x,app_2(l,k))
r31: sum_1(cons_2(x,nil_0())) -> cons_2(x,nil_0())
r32: sum_1(cons_2(x,cons_2(y,l))) -> sum_1(cons_2(plus_2(x,y),l))
r33: sum_1(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
r34: map_2(f,nil_0()) -> nil_0()
r35: map_2(f,cons_2(x,xs)) -> cons_2(|app'|(f,x),map_2(f,xs))
r36: filter_2(f,nil_0()) -> nil_0()
r37: filter_2(f,cons_2(x,xs)) -> filter2_4(|app'|(f,x),f,x,xs)
r38: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r39: 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:
          
            app_2#_A(x1,x2) = max{0, x1 - 30}
            cons_2_A(x1,x2) = max{x1 - 134, x2 + 29}
            app_2#!_A(x1,x2) = max{10, x1 - 30, x2 - 1}
            cons_2!_A(x1,x2) = max{9, x1 - 2182, x2 - 20}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app_2#_A(x1,x2) = max{0, x1 - 7}
            cons_2_A(x1,x2) = max{x1 - 50, x2 + 6}
            app_2#!_A(x1,x2) = max{7, x1 - 7, x2 - 1}
            cons_2!_A(x1,x2) = max{6, x1 - 152, x2 - 40}
        
  
    partial status:
  
      pi(app_2#) = [1]
      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: quot_2#(s_1(x),s_1(y)) -> quot_2#(minus_2(x,y),s_1(y))

and R consists of:

r1: |app'|(app_0(),y) -> app_1(y)
r2: |app'|(app_1(x1),y) -> app_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'|(minus_0(),y) -> minus_1(y)
r14: |app'|(minus_1(x1),y) -> minus_2(x1,y)
r15: |app'|(plus_0(),y) -> plus_1(y)
r16: |app'|(plus_1(x1),y) -> plus_2(x1,y)
r17: |app'|(quot_0(),y) -> quot_1(y)
r18: |app'|(quot_1(x1),y) -> quot_2(x1,y)
r19: |app'|(s_0(),y) -> s_1(y)
r20: |app'|(sum_0(),y) -> sum_1(y)
r21: minus_2(x,|0|_0()) -> x
r22: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r23: minus_2(minus_2(x,y),z) -> minus_2(x,plus_2(y,z))
r24: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r25: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r26: plus_2(|0|_0(),y) -> y
r27: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r28: app_2(nil_0(),k) -> k
r29: app_2(l,nil_0()) -> l
r30: app_2(cons_2(x,l),k) -> cons_2(x,app_2(l,k))
r31: sum_1(cons_2(x,nil_0())) -> cons_2(x,nil_0())
r32: sum_1(cons_2(x,cons_2(y,l))) -> sum_1(cons_2(plus_2(x,y),l))
r33: sum_1(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
r34: map_2(f,nil_0()) -> nil_0()
r35: map_2(f,cons_2(x,xs)) -> cons_2(|app'|(f,x),map_2(f,xs))
r36: filter_2(f,nil_0()) -> nil_0()
r37: filter_2(f,cons_2(x,xs)) -> filter2_4(|app'|(f,x),f,x,xs)
r38: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r39: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r21, r22, r23, r26, r27

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:
          
            quot_2#_A(x1,x2) = max{0, x1 - 21, x2 - 34}
            s_1_A(x1) = max{19, x1}
            minus_2_A(x1,x2) = max{20, x1, x2 - 13}
            |0|_0_A = 36
            plus_2_A(x1,x2) = max{30, x1 - 20, x2}
            quot_2#!_A(x1,x2) = max{4, x1 - 38, x2 - 50}
            s_1!_A(x1) = max{4, x1 - 51}
            minus_2!_A(x1,x2) = 0
            |0|_0!_A = 2
            plus_2!_A(x1,x2) = max{5, x1 - 20, x2 - 20}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot_2#_A(x1,x2) = max{0, x1 - 6, x2 - 6}
            s_1_A(x1) = max{4, x1}
            minus_2_A(x1,x2) = max{5, x1, x2 - 58}
            |0|_0_A = 4
            plus_2_A(x1,x2) = max{57, x1 - 7, x2}
            quot_2#!_A(x1,x2) = max{49, x1 - 9, x2 - 66}
            s_1!_A(x1) = max{50, x1 - 65}
            minus_2!_A(x1,x2) = max{48, x1 - 64, x2 - 115}
            |0|_0!_A = 0
            plus_2!_A(x1,x2) = max{49, x1 - 13, x2 - 67}
        
  
    partial status:
  
      pi(quot_2#) = [1]
      pi(s_1) = [1]
      pi(minus_2) = [1]
      pi(|0|_0) = []
      pi(plus_2) = [1, 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: minus_2#(s_1(x),s_1(y)) -> minus_2#(x,y)
p2: minus_2#(minus_2(x,y),z) -> minus_2#(x,plus_2(y,z))

and R consists of:

r1: |app'|(app_0(),y) -> app_1(y)
r2: |app'|(app_1(x1),y) -> app_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'|(minus_0(),y) -> minus_1(y)
r14: |app'|(minus_1(x1),y) -> minus_2(x1,y)
r15: |app'|(plus_0(),y) -> plus_1(y)
r16: |app'|(plus_1(x1),y) -> plus_2(x1,y)
r17: |app'|(quot_0(),y) -> quot_1(y)
r18: |app'|(quot_1(x1),y) -> quot_2(x1,y)
r19: |app'|(s_0(),y) -> s_1(y)
r20: |app'|(sum_0(),y) -> sum_1(y)
r21: minus_2(x,|0|_0()) -> x
r22: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r23: minus_2(minus_2(x,y),z) -> minus_2(x,plus_2(y,z))
r24: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r25: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r26: plus_2(|0|_0(),y) -> y
r27: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r28: app_2(nil_0(),k) -> k
r29: app_2(l,nil_0()) -> l
r30: app_2(cons_2(x,l),k) -> cons_2(x,app_2(l,k))
r31: sum_1(cons_2(x,nil_0())) -> cons_2(x,nil_0())
r32: sum_1(cons_2(x,cons_2(y,l))) -> sum_1(cons_2(plus_2(x,y),l))
r33: sum_1(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
r34: map_2(f,nil_0()) -> nil_0()
r35: map_2(f,cons_2(x,xs)) -> cons_2(|app'|(f,x),map_2(f,xs))
r36: filter_2(f,nil_0()) -> nil_0()
r37: filter_2(f,cons_2(x,xs)) -> filter2_4(|app'|(f,x),f,x,xs)
r38: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r39: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r21, r26, r27

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:
          
            minus_2#_A(x1,x2) = max{0, x1 - 22, x2 - 48}
            s_1_A(x1) = max{18, x1}
            minus_2_A(x1,x2) = max{20, x1 + 1, x2 - 49}
            plus_2_A(x1,x2) = max{47, x1 - 23, x2}
            |0|_0_A = 18
            minus_2#!_A(x1,x2) = max{7, x1 - 21, x2 - 95}
            s_1!_A(x1) = max{4, x1 - 72}
            minus_2!_A(x1,x2) = max{6, x1 - 47, x2 - 72}
            plus_2!_A(x1,x2) = max{5, x1 - 71, x2 - 24}
            |0|_0!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus_2#_A(x1,x2) = max{0, x1 - 30, x2 - 55}
            s_1_A(x1) = max{27, x1}
            minus_2_A(x1,x2) = max{28, x1 + 1, x2 - 60}
            plus_2_A(x1,x2) = max{54, x1 - 56, x2}
            |0|_0_A = 28
            minus_2#!_A(x1,x2) = max{6, x1 - 89, x2 - 204}
            s_1!_A(x1) = max{5, x1 - 91}
            minus_2!_A(x1,x2) = max{3, x1 - 90, x2 - 261}
            plus_2!_A(x1,x2) = max{4, x1 - 148, x2 - 203}
            |0|_0!_A = 0
        
  
    partial status:
  
      pi(minus_2#) = [1]
      pi(s_1) = [1]
      pi(minus_2) = [1]
      pi(plus_2) = [1, 2]
      pi(|0|_0) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- Reduction pair.

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

p1: sum_1#(cons_2(x,cons_2(y,l))) -> sum_1#(cons_2(plus_2(x,y),l))

and R consists of:

r1: |app'|(app_0(),y) -> app_1(y)
r2: |app'|(app_1(x1),y) -> app_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'|(minus_0(),y) -> minus_1(y)
r14: |app'|(minus_1(x1),y) -> minus_2(x1,y)
r15: |app'|(plus_0(),y) -> plus_1(y)
r16: |app'|(plus_1(x1),y) -> plus_2(x1,y)
r17: |app'|(quot_0(),y) -> quot_1(y)
r18: |app'|(quot_1(x1),y) -> quot_2(x1,y)
r19: |app'|(s_0(),y) -> s_1(y)
r20: |app'|(sum_0(),y) -> sum_1(y)
r21: minus_2(x,|0|_0()) -> x
r22: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r23: minus_2(minus_2(x,y),z) -> minus_2(x,plus_2(y,z))
r24: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r25: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r26: plus_2(|0|_0(),y) -> y
r27: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r28: app_2(nil_0(),k) -> k
r29: app_2(l,nil_0()) -> l
r30: app_2(cons_2(x,l),k) -> cons_2(x,app_2(l,k))
r31: sum_1(cons_2(x,nil_0())) -> cons_2(x,nil_0())
r32: sum_1(cons_2(x,cons_2(y,l))) -> sum_1(cons_2(plus_2(x,y),l))
r33: sum_1(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
r34: map_2(f,nil_0()) -> nil_0()
r35: map_2(f,cons_2(x,xs)) -> cons_2(|app'|(f,x),map_2(f,xs))
r36: filter_2(f,nil_0()) -> nil_0()
r37: filter_2(f,cons_2(x,xs)) -> filter2_4(|app'|(f,x),f,x,xs)
r38: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r39: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r26, r27

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:
          
            sum_1#_A(x1) = 0
            cons_2_A(x1,x2) = 0
            plus_2_A(x1,x2) = max{13, x1 - 26, x2 + 2}
            |0|_0_A = 9
            s_1_A(x1) = max{13, x1 - 14}
            sum_1#!_A(x1) = max{10, x1 - 1}
            cons_2!_A(x1,x2) = x2 + 61
            plus_2!_A(x1,x2) = max{x1 - 14, x2 + 36}
            |0|_0!_A = 0
            s_1!_A(x1) = max{4, x1 - 3}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum_1#_A(x1) = max{1, x1 - 31}
            cons_2_A(x1,x2) = max{x1 - 10, x2 + 15}
            plus_2_A(x1,x2) = max{42, x1 - 238, x2 + 14}
            |0|_0_A = 452
            s_1_A(x1) = max{16, x1 - 43}
            sum_1#!_A(x1) = 1
            cons_2!_A(x1,x2) = 0
            plus_2!_A(x1,x2) = max{300, x1 - 17, x2 + 252}
            |0|_0!_A = 302
            s_1!_A(x1) = 301
        
  
    partial status:
  
      pi(sum_1#) = [1]
      pi(cons_2) = [2]
      pi(plus_2) = [1, 2]
      pi(|0|_0) = []
      pi(s_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#(s_1(x),y) -> plus_2#(x,y)

and R consists of:

r1: |app'|(app_0(),y) -> app_1(y)
r2: |app'|(app_1(x1),y) -> app_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'|(minus_0(),y) -> minus_1(y)
r14: |app'|(minus_1(x1),y) -> minus_2(x1,y)
r15: |app'|(plus_0(),y) -> plus_1(y)
r16: |app'|(plus_1(x1),y) -> plus_2(x1,y)
r17: |app'|(quot_0(),y) -> quot_1(y)
r18: |app'|(quot_1(x1),y) -> quot_2(x1,y)
r19: |app'|(s_0(),y) -> s_1(y)
r20: |app'|(sum_0(),y) -> sum_1(y)
r21: minus_2(x,|0|_0()) -> x
r22: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r23: minus_2(minus_2(x,y),z) -> minus_2(x,plus_2(y,z))
r24: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r25: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r26: plus_2(|0|_0(),y) -> y
r27: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r28: app_2(nil_0(),k) -> k
r29: app_2(l,nil_0()) -> l
r30: app_2(cons_2(x,l),k) -> cons_2(x,app_2(l,k))
r31: sum_1(cons_2(x,nil_0())) -> cons_2(x,nil_0())
r32: sum_1(cons_2(x,cons_2(y,l))) -> sum_1(cons_2(plus_2(x,y),l))
r33: sum_1(app_2(l,cons_2(x,cons_2(y,k)))) -> sum_1(app_2(l,sum_1(cons_2(x,cons_2(y,k)))))
r34: map_2(f,nil_0()) -> nil_0()
r35: map_2(f,cons_2(x,xs)) -> cons_2(|app'|(f,x),map_2(f,xs))
r36: filter_2(f,nil_0()) -> nil_0()
r37: filter_2(f,cons_2(x,xs)) -> filter2_4(|app'|(f,x),f,x,xs)
r38: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r39: 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) = max{0, x1 - 4}
            s_1_A(x1) = max{148, x1 + 3}
            plus_2#!_A(x1,x2) = max{3, x1 - 4}
            s_1!_A(x1) = max{2, x1 - 305}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = max{0, x1 - 14}
            s_1_A(x1) = max{17, x1 + 13}
            plus_2#!_A(x1,x2) = max{9, x1 - 39, x2 - 1}
            s_1!_A(x1) = max{8, x1 - 40}
        
  
    partial status:
  
      pi(plus_2#) = []
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.