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

  (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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = x1
              filter_1_A(x1) = x1 + 2
              filter_2#_A(x1,x2) = x1
              cons_2_A(x1,x2) = x1 + x2 + 5
              map_1_A(x1) = x1 + 2
              map_2#_A(x1,x2) = x1
              filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 2
              filter2_4#_A(x1,x2,x3,x4) = x2
              false_0_A() = 1
              app_A(x1,x2) = x2 + 1
              true_0_A() = 0
              app#!_A(x1,x2) = x2 + 7
              filter_1!_A(x1) = 5
              filter_2#!_A(x1,x2) = x2 + 3
              cons_2!_A(x1,x2) = x2 + 5
              map_1!_A(x1) = 2
              map_2#!_A(x1,x2) = x2 + 4
              filter2_3!_A(x1,x2,x3) = 3
              filter2_4#!_A(x1,x2,x3,x4) = x4 + 6
              false_0!_A() = 7
              app!_A(x1,x2) = 4
              true_0!_A() = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 1
            filter_1_A(x1) = max{3, x1 - 8}
            filter_2#_A(x1,x2) = 1
            cons_2_A(x1,x2) = max{22, x1 + 21}
            map_1_A(x1) = max{3, x1 - 8}
            map_2#_A(x1,x2) = 1
            filter2_3_A(x1,x2,x3) = max{3, x1 - 6, x2 - 10, x3 - 6}
            filter2_4#_A(x1,x2,x3,x4) = 1
            false_0_A = 0
            app_A(x1,x2) = max{2, x1 - 15, x2 - 15}
            true_0_A = 4
            app#!_A(x1,x2) = 11
            filter_1!_A(x1) = max{0, x1 - 20}
            filter_2#!_A(x1,x2) = max{12, x1 + 10, x2 - 17}
            cons_2!_A(x1,x2) = max{x1 + 14, x2 + 21}
            map_1!_A(x1) = 12
            map_2#!_A(x1,x2) = 12
            filter2_3!_A(x1,x2,x3) = max{2, x1 - 2, x2 - 2, x3 - 2}
            filter2_4#!_A(x1,x2,x3,x4) = 3
            false_0!_A = 28
            app!_A(x1,x2) = 13
            true_0!_A = 26
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(map_1) = []
      pi(map_2#) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = []
      pi(true_0) = []

The following rules are removed:

  p1, p2, p3, p4, p5, p6, p7, p8, p9

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              times_2#_A(x1,x2) = x2
              s_1_A(x1) = x1 + 3
              times_2#!_A(x1,x2) = x2 + 4
              s_1!_A(x1) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times_2#_A(x1,x2) = x2
            s_1_A(x1) = x1 + 5
            times_2#!_A(x1,x2) = max{8, x2 - 6}
            s_1!_A(x1) = max{7, x1 - 14}
        
  
    partial status:
  
      pi(times_2#) = []
      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#(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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus_2#_A(x1,x2) = x1 + x2
              s_1_A(x1) = x1 + 8
              plus_2#!_A(x1,x2) = 9
              s_1!_A(x1) = 9
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = max{x1, x2}
            s_1_A(x1) = x1 + 14
            plus_2#!_A(x1,x2) = max{10, x2 - 1}
            s_1!_A(x1) = max{10, x1 + 9}
        
  
    partial status:
  
      pi(plus_2#) = []
      pi(s_1) = [1]

The following rules are removed:

  p1



-- SCC decomposition.

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(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#(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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus_2#_A(x1,x2) = x1
              s_1_A(x1) = x1 + 3
              plus_2#!_A(x1,x2) = x1 + x2
              s_1!_A(x1) = x1 + 14
        
        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{10, x1 - 11}
            plus_2#!_A(x1,x2) = max{18, x1 - 11, x2 - 1}
            s_1!_A(x1) = x1 + 17
        
  
    partial status:
  
      pi(plus_2#) = []
      pi(s_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.