YES
We show the termination of the TRS R:

  app(f(),|0|()) -> true()
  app(f(),|1|()) -> false()
  app(f(),app(s(),x)) -> app(f(),x)
  app(app(app(if(),true()),app(s(),x)),app(s(),y)) -> app(s(),x)
  app(app(app(if(),false()),app(s(),x)),app(s(),y)) -> app(s(),y)
  app(app(g(),x),app(c(),y)) -> app(c(),app(app(g(),x),y))
  app(app(g(),x),app(c(),y)) -> app(app(g(),x),app(app(app(if(),app(f(),x)),app(c(),app(app(g(),app(s(),x)),y))),app(c(),y)))
  app(app(map(),|fun|),nil()) -> nil()
  app(app(map(),|fun|),app(app(cons(),x),xs)) -> app(app(cons(),app(|fun|,x)),app(app(map(),|fun|),xs))
  app(app(filter(),|fun|),nil()) -> nil()
  app(app(filter(),|fun|),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(|fun|,x)),|fun|),x),xs)
  app(app(app(app(filter2(),true()),|fun|),x),xs) -> app(app(cons(),x),app(app(filter(),|fun|),xs))
  app(app(app(app(filter2(),false()),|fun|),x),xs) -> app(app(filter(),|fun|),xs)

Instead we prove termination of the uncurried system:

  app(c_0(),y) -> c_1(y)
  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(f_0(),y) -> f_1(y)
  app(filter_0(),y) -> filter_1(y)
  app(filter_1(x1),y) -> filter_2(x1,y)
  app(filter2_0(),y) -> filter2_1(y)
  app(filter2_1(x1),y) -> filter2_2(x1,y)
  app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
  app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
  app(g_0(),y) -> g_1(y)
  app(g_1(x1),y) -> g_2(x1,y)
  app(if_0(),y) -> if_1(y)
  app(if_1(x1),y) -> if_2(x1,y)
  app(if_2(x1,x2),y) -> if_3(x1,x2,y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(s_0(),y) -> s_1(y)
  f_1(|0|_0()) -> true_0()
  f_1(|1|_0()) -> false_0()
  f_1(s_1(x)) -> f_1(x)
  if_3(true_0(),s_1(x),s_1(y)) -> s_1(x)
  if_3(false_0(),s_1(x),s_1(y)) -> s_1(y)
  g_2(x,c_1(y)) -> c_1(g_2(x,y))
  g_2(x,c_1(y)) -> g_2(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y)))
  map_2(|fun|,nil_0()) -> nil_0()
  map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
  filter_2(|fun|,nil_0()) -> nil_0()
  filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
  filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
  filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)


-- SCC decomposition.

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

p1: app#(f_0(),y) -> f_1#(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#(g_1(x1),y) -> g_2#(x1,y)
p5: app#(if_2(x1,x2),y) -> if_3#(x1,x2,y)
p6: app#(map_1(x1),y) -> map_2#(x1,y)
p7: f_1#(s_1(x)) -> f_1#(x)
p8: g_2#(x,c_1(y)) -> g_2#(x,y)
p9: g_2#(x,c_1(y)) -> g_2#(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y)))
p10: g_2#(x,c_1(y)) -> if_3#(f_1(x),c_1(g_2(s_1(x),y)),c_1(y))
p11: g_2#(x,c_1(y)) -> f_1#(x)
p12: g_2#(x,c_1(y)) -> g_2#(s_1(x),y)
p13: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p14: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p15: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p16: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p17: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p18: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)

and R consists of:

r1: app(c_0(),y) -> c_1(y)
r2: app(cons_0(),y) -> cons_1(y)
r3: app(cons_1(x1),y) -> cons_2(x1,y)
r4: app(f_0(),y) -> f_1(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(g_0(),y) -> g_1(y)
r12: app(g_1(x1),y) -> g_2(x1,y)
r13: app(if_0(),y) -> if_1(y)
r14: app(if_1(x1),y) -> if_2(x1,y)
r15: app(if_2(x1,x2),y) -> if_3(x1,x2,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: app(s_0(),y) -> s_1(y)
r19: f_1(|0|_0()) -> true_0()
r20: f_1(|1|_0()) -> false_0()
r21: f_1(s_1(x)) -> f_1(x)
r22: if_3(true_0(),s_1(x),s_1(y)) -> s_1(x)
r23: if_3(false_0(),s_1(x),s_1(y)) -> s_1(y)
r24: g_2(x,c_1(y)) -> c_1(g_2(x,y))
r25: g_2(x,c_1(y)) -> g_2(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y)))
r26: map_2(|fun|,nil_0()) -> nil_0()
r27: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r28: filter_2(|fun|,nil_0()) -> nil_0()
r29: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r30: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r31: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p6, p13, p14, p15, p16, p17, p18}
  {p8, p12}
  {p7}


-- Reduction pair.

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

p1: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p4: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p5: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p6: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p7: app#(filter_1(x1),y) -> filter_2#(x1,y)
p8: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p9: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)

and R consists of:

r1: app(c_0(),y) -> c_1(y)
r2: app(cons_0(),y) -> cons_1(y)
r3: app(cons_1(x1),y) -> cons_2(x1,y)
r4: app(f_0(),y) -> f_1(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(g_0(),y) -> g_1(y)
r12: app(g_1(x1),y) -> g_2(x1,y)
r13: app(if_0(),y) -> if_1(y)
r14: app(if_1(x1),y) -> if_2(x1,y)
r15: app(if_2(x1,x2),y) -> if_3(x1,x2,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: app(s_0(),y) -> s_1(y)
r19: f_1(|0|_0()) -> true_0()
r20: f_1(|1|_0()) -> false_0()
r21: f_1(s_1(x)) -> f_1(x)
r22: if_3(true_0(),s_1(x),s_1(y)) -> s_1(x)
r23: if_3(false_0(),s_1(x),s_1(y)) -> s_1(y)
r24: g_2(x,c_1(y)) -> c_1(g_2(x,y))
r25: g_2(x,c_1(y)) -> g_2(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y)))
r26: map_2(|fun|,nil_0()) -> nil_0()
r27: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r28: filter_2(|fun|,nil_0()) -> nil_0()
r29: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r30: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r31: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r1, r13, r18, r22, r23, r26

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{3, x1 - 4, x2 - 3}
            cons_2_A(x1,x2) = max{x1 + 4, x2 + 5}
            app#_A(x1,x2) = max{3, x1 - 16, x2 - 1}
            map_1_A(x1) = x1 + 12
            filter2_3_A(x1,x2,x3) = max{17, x2 + 12, x3 + 12}
            filter2_4#_A(x1,x2,x3,x4) = max{3, x2 - 11, x3 - 4, x4 - 1}
            false_0_A = 12
            filter_2#_A(x1,x2) = max{3, x1 - 11, x2 - 3}
            filter_1_A(x1) = x1 + 5
            app_A(x1,x2) = 12
            true_0_A = 12
            c_0_A = 12
            c_1_A(x1) = 0
            if_0_A = 37
            if_1_A(x1) = 12
            s_0_A = 12
            s_1_A(x1) = 1
            if_3_A(x1,x2,x3) = 3
            map_2_A(x1,x2) = 3
            nil_0_A = 2
            map_2#!_A(x1,x2) = max{x1 + 12, x2 - 6}
            cons_2!_A(x1,x2) = max{3, x2 + 1}
            app#!_A(x1,x2) = max{2, x1, x2 - 3}
            map_1!_A(x1) = max{1, x1 - 18}
            filter2_3!_A(x1,x2,x3) = max{3, x1 + 1}
            filter2_4#!_A(x1,x2,x3,x4) = max{x2 + 5, x3 - 1, x4 - 4}
            false_0!_A = 15
            filter_2#!_A(x1,x2) = max{x1 + 5, x2 - 5}
            filter_1!_A(x1) = max{6, x1 - 1}
            app!_A(x1,x2) = max{18, x1 - 16, x2 + 16}
            true_0!_A = 8
            c_0!_A = 2
            c_1!_A(x1) = x1 + 14
            if_0!_A = 21
            if_1!_A(x1) = 20
            s_0!_A = 13
            s_1!_A(x1) = 14
            if_3!_A(x1,x2,x3) = max{20, x1 - 13, x3 - 2}
            map_2!_A(x1,x2) = max{9, x1 - 17, x2 - 5}
            nil_0!_A = 8
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 15, x2 - 19}
            cons_2_A(x1,x2) = max{x1 + 18, x2 + 5}
            app#_A(x1,x2) = max{1, x1 - 16, x2 - 18}
            map_1_A(x1) = max{11, x1 + 1}
            filter2_3_A(x1,x2,x3) = max{11, x1 - 39, x2 + 1, x3 + 1}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 15, x3 - 15, x4 - 18}
            false_0_A = 11
            filter_2#_A(x1,x2) = max{1, x1 - 15, x2 - 18}
            filter_1_A(x1) = max{15, x1 + 1}
            app_A(x1,x2) = max{45, x1 - 19, x2 + 11}
            true_0_A = 11
            c_0_A = 13
            c_1_A(x1) = max{0, x1 - 1}
            if_0_A = 13
            if_1_A(x1) = max{15, x1 - 1}
            s_0_A = 13
            s_1_A(x1) = x1 + 8
            if_3_A(x1,x2,x3) = max{x1 - 39, x2 + 6, x3 + 7}
            map_2_A(x1,x2) = max{16, x1 + 6, x2 + 11}
            nil_0_A = 20
            map_2#!_A(x1,x2) = x2 + 1
            cons_2!_A(x1,x2) = 0
            app#!_A(x1,x2) = x2 + 3
            map_1!_A(x1) = max{0, x1 - 16}
            filter2_3!_A(x1,x2,x3) = 0
            filter2_4#!_A(x1,x2,x3,x4) = max{2, x4 - 1}
            false_0!_A = 4
            filter_2#!_A(x1,x2) = max{1, x2 - 2}
            filter_1!_A(x1) = max{0, x1 - 1}
            app!_A(x1,x2) = max{24, x1 - 14, x2 + 21}
            true_0!_A = 2
            c_0!_A = 1
            c_1!_A(x1) = max{0, x1 - 13}
            if_0!_A = 2
            if_1!_A(x1) = 1
            s_0!_A = 23
            s_1!_A(x1) = max{22, x1 - 10}
            if_3!_A(x1,x2,x3) = max{26, x1 - 46, x2 - 9, x3 - 11}
            map_2!_A(x1,x2) = max{0, x1 - 15, x2 - 21}
            nil_0!_A = 0
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app#) = []
      pi(map_1) = [1]
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(filter_1) = [1]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(c_0) = []
      pi(c_1) = []
      pi(if_0) = []
      pi(if_1) = [1]
      pi(s_0) = []
      pi(s_1) = []
      pi(if_3) = [1, 3]
      pi(map_2) = [1, 2]
      pi(nil_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: g_2#(x,c_1(y)) -> g_2#(s_1(x),y)
p2: g_2#(x,c_1(y)) -> g_2#(x,y)

and R consists of:

r1: app(c_0(),y) -> c_1(y)
r2: app(cons_0(),y) -> cons_1(y)
r3: app(cons_1(x1),y) -> cons_2(x1,y)
r4: app(f_0(),y) -> f_1(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(g_0(),y) -> g_1(y)
r12: app(g_1(x1),y) -> g_2(x1,y)
r13: app(if_0(),y) -> if_1(y)
r14: app(if_1(x1),y) -> if_2(x1,y)
r15: app(if_2(x1,x2),y) -> if_3(x1,x2,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: app(s_0(),y) -> s_1(y)
r19: f_1(|0|_0()) -> true_0()
r20: f_1(|1|_0()) -> false_0()
r21: f_1(s_1(x)) -> f_1(x)
r22: if_3(true_0(),s_1(x),s_1(y)) -> s_1(x)
r23: if_3(false_0(),s_1(x),s_1(y)) -> s_1(y)
r24: g_2(x,c_1(y)) -> c_1(g_2(x,y))
r25: g_2(x,c_1(y)) -> g_2(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y)))
r26: map_2(|fun|,nil_0()) -> nil_0()
r27: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r28: filter_2(|fun|,nil_0()) -> nil_0()
r29: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r30: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r31: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r19, r20

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:
          
            g_2#_A(x1,x2) = max{0, x1 - 6, x2 - 17}
            c_1_A(x1) = max{15, x1 + 8}
            s_1_A(x1) = max{5, x1 - 1}
            f_1_A(x1) = max{21, x1 - 1}
            |0|_0_A = 0
            true_0_A = 0
            |1|_0_A = 0
            false_0_A = 0
            g_2#!_A(x1,x2) = max{8, x1 - 6, x2 - 80}
            c_1!_A(x1) = max{7, x1 - 81}
            s_1!_A(x1) = max{4, x1 - 89}
            f_1!_A(x1) = max{2, x1 - 89}
            |0|_0!_A = 0
            true_0!_A = 1
            |1|_0!_A = 0
            false_0!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g_2#_A(x1,x2) = max{0, x1 - 4, x2 - 20}
            c_1_A(x1) = max{18, x1 + 13}
            s_1_A(x1) = max{3, x1 - 1}
            f_1_A(x1) = max{20, x1 - 2}
            |0|_0_A = 0
            true_0_A = 0
            |1|_0_A = 0
            false_0_A = 0
            g_2#!_A(x1,x2) = max{8, x1 - 4, x2 - 51}
            c_1!_A(x1) = max{0, x1 - 50}
            s_1!_A(x1) = max{7, x1 - 81}
            f_1!_A(x1) = max{7, x1 - 81}
            |0|_0!_A = 0
            true_0!_A = 6
            |1|_0!_A = 7
            false_0!_A = 6
        
  
    partial status:
  
      pi(g_2#) = [2]
      pi(c_1) = [1]
      pi(s_1) = []
      pi(f_1) = []
      pi(|0|_0) = []
      pi(true_0) = []
      pi(|1|_0) = []
      pi(false_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: f_1#(s_1(x)) -> f_1#(x)

and R consists of:

r1: app(c_0(),y) -> c_1(y)
r2: app(cons_0(),y) -> cons_1(y)
r3: app(cons_1(x1),y) -> cons_2(x1,y)
r4: app(f_0(),y) -> f_1(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(g_0(),y) -> g_1(y)
r12: app(g_1(x1),y) -> g_2(x1,y)
r13: app(if_0(),y) -> if_1(y)
r14: app(if_1(x1),y) -> if_2(x1,y)
r15: app(if_2(x1,x2),y) -> if_3(x1,x2,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: app(s_0(),y) -> s_1(y)
r19: f_1(|0|_0()) -> true_0()
r20: f_1(|1|_0()) -> false_0()
r21: f_1(s_1(x)) -> f_1(x)
r22: if_3(true_0(),s_1(x),s_1(y)) -> s_1(x)
r23: if_3(false_0(),s_1(x),s_1(y)) -> s_1(y)
r24: g_2(x,c_1(y)) -> c_1(g_2(x,y))
r25: g_2(x,c_1(y)) -> g_2(x,if_3(f_1(x),c_1(g_2(s_1(x),y)),c_1(y)))
r26: map_2(|fun|,nil_0()) -> nil_0()
r27: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r28: filter_2(|fun|,nil_0()) -> nil_0()
r29: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r30: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r31: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r19, r20

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:
          
            f_1#_A(x1) = x1
            s_1_A(x1) = max{3, x1 + 1}
            f_1_A(x1) = max{16, x1 - 49}
            |0|_0_A = 48
            true_0_A = 6
            |1|_0_A = 48
            false_0_A = 0
            f_1#!_A(x1) = max{4, x1 - 47}
            s_1!_A(x1) = max{4, x1 - 46}
            f_1!_A(x1) = max{3, x1 - 45}
            |0|_0!_A = 3
            true_0!_A = 2
            |1|_0!_A = 3
            false_0!_A = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f_1#_A(x1) = x1
            s_1_A(x1) = x1 + 17
            f_1_A(x1) = max{22, x1 - 7}
            |0|_0_A = 0
            true_0_A = 0
            |1|_0_A = 6
            false_0_A = 0
            f_1#!_A(x1) = max{1, x1 - 87}
            s_1!_A(x1) = max{8, x1 - 87}
            f_1!_A(x1) = max{4, x1 - 86}
            |0|_0!_A = 6
            true_0!_A = 5
            |1|_0!_A = 4
            false_0!_A = 3
        
  
    partial status:
  
      pi(f_1#) = [1]
      pi(s_1) = [1]
      pi(f_1) = []
      pi(|0|_0) = []
      pi(true_0) = []
      pi(|1|_0) = []
      pi(false_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.