YES
We show the termination of the TRS R:

  app(app(times(),x),app(app(plus(),y),app(s(),z))) -> app(app(plus(),app(app(times(),x),app(app(plus(),y),app(app(times(),app(s(),z)),|0|())))),app(app(times(),x),app(s(),z)))
  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(),x),app(s(),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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
  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(x,s_1(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,plus_2(y,s_1(z))) -> plus_2#(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
p7: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,plus_2(y,times_2(s_1(z),|0|_0())))
p8: times_2#(x,plus_2(y,s_1(z))) -> plus_2#(y,times_2(s_1(z),|0|_0()))
p9: times_2#(x,plus_2(y,s_1(z))) -> times_2#(s_1(z),|0|_0())
p10: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,s_1(z))
p11: times_2#(x,s_1(y)) -> plus_2#(times_2(x,y),x)
p12: times_2#(x,s_1(y)) -> times_2#(x,y)
p13: plus_2#(x,s_1(y)) -> plus_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(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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p14, p15, p16, p17, p18, p19}
  {p7, p10, p12}
  {p13}


-- 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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{3, x1 + 1}
        filter_1_A(x1) = x1 + 26
        filter_2#_A(x1,x2) = max{6, x1 + 2}
        cons_2_A(x1,x2) = 5
        map_1_A(x1) = x1 + 33
        map_2#_A(x1,x2) = max{18, x1 + 2}
        filter2_3_A(x1,x2,x3) = max{20, x1 - 2, x2 + 18, x3 - 2}
        filter2_4#_A(x1,x2,x3,x4) = max{6, x2 + 2}
        false_0_A = 5
        app_A(x1,x2) = max{11, x1 - 17, x2 + 4}
        true_0_A = 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{1, x2 - 3}
          filter_1_A(x1) = max{30, x1 - 52}
          filter_2#_A(x1,x2) = max{1, x2 - 127}
          cons_2_A(x1,x2) = max{127, x1 + 125, x2}
          map_1_A(x1) = x1 + 17
          map_2#_A(x1,x2) = max{1, x2 - 128}
          filter2_3_A(x1,x2,x3) = max{x1 + 34, x2 + 18, x3 + 30}
          filter2_4#_A(x1,x2,x3,x4) = max{1, x4 - 127}
          false_0_A = 0
          app_A(x1,x2) = max{29, x1 - 18}
          true_0_A = 109
          app#!_A(x1,x2) = max{45, x1 - 31, x2 - 25}
          filter_1!_A(x1) = max{41, x1 - 2}
          filter_2#!_A(x1,x2) = max{44, x1 - 84, x2 - 84}
          cons_2!_A(x1,x2) = max{40, x1 - 7}
          map_1!_A(x1) = max{41, x1 + 7}
          map_2#!_A(x1,x2) = max{42, x1 - 32, x2 - 151}
          filter2_3!_A(x1,x2,x3) = max{2, x1 - 68, x2 - 84, x3 - 7}
          filter2_4#!_A(x1,x2,x3,x4) = max{44, x2 - 84, x3 - 8, x4 - 84}
          false_0!_A = 1
          app!_A(x1,x2) = max{40, x1 - 103, x2 - 3}
          true_0!_A = 1
  
    partial status:
  
      pi(app#) = [1]
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(filter2_3) = [2]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [2]
      pi(true_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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2, p3, 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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = x1
        cons_2_A(x1,x2) = max{4, x1 - 10, x2 - 11}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          map_2#_A(x1,x2) = max{0, x2 - 13}
          cons_2_A(x1,x2) = max{18, x1 - 77, x2 + 12}
          map_2#!_A(x1,x2) = max{4, x2 - 13}
          cons_2!_A(x1,x2) = max{3, x1 - 108, x2 - 20}
  
    partial status:
  
      pi(map_2#) = [1]
      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(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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter2_4#_A(x1,x2,x3,x4) = max{5, x4 + 3}
        false_0_A = 33
        filter_2#_A(x1,x2) = x2 + 2
        cons_2_A(x1,x2) = max{10, x2 + 4}
        app_A(x1,x2) = max{x1 + 13, x2 + 17}
        true_0_A = 37
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          filter2_4#_A(x1,x2,x3,x4) = 1
          false_0_A = 0
          filter_2#_A(x1,x2) = 1
          cons_2_A(x1,x2) = max{31, x1 + 29}
          app_A(x1,x2) = max{7, x1 - 28}
          true_0_A = 2
          filter2_4#!_A(x1,x2,x3,x4) = 16
          false_0!_A = 2
          filter_2#!_A(x1,x2) = max{16, x2 - 16}
          cons_2!_A(x1,x2) = max{17, x1 + 12, x2 + 2}
          app!_A(x1,x2) = max{24, x1 - 9, x2 + 22}
          true_0!_A = 4
  
    partial status:
  
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = []
      pi(app) = [1]
      pi(true_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)
p2: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,s_1(z))
p3: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,plus_2(y,times_2(s_1(z),|0|_0())))

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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r17, r19, r20

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times_2#_A(x1,x2) = max{11, x2}
        s_1_A(x1) = max{11, x1}
        plus_2_A(x1,x2) = max{x1 + 39, x2 + 12}
        times_2_A(x1,x2) = max{22, x1 - 12, x2 + 4}
        |0|_0_A = 10
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          times_2#_A(x1,x2) = max{0, x1 - 18, x2 - 17}
          s_1_A(x1) = max{10, x1}
          plus_2_A(x1,x2) = max{x1 + 6, x2 + 6}
          times_2_A(x1,x2) = max{0, x1 - 16, x2 - 6}
          |0|_0_A = 0
          times_2#!_A(x1,x2) = 0
          s_1!_A(x1) = 0
          plus_2!_A(x1,x2) = x2 + 5
          times_2!_A(x1,x2) = max{3, x1 - 12, x2 - 9}
          |0|_0!_A = 2
  
    partial status:
  
      pi(times_2#) = [2]
      pi(s_1) = [1]
      pi(plus_2) = [2]
      pi(times_2) = [2]
      pi(|0|_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: 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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus_2#_A(x1,x2) = x1 + 1
        s_1_A(x1) = max{0, x1 - 2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus_2#_A(x1,x2) = max{0, x2 - 2}
          s_1_A(x1) = max{7, x1 + 1}
          plus_2#!_A(x1,x2) = max{1, x2 - 2}
          s_1!_A(x1) = max{0, x1 - 1}
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.