YES
We show the termination of the TRS R:

  app(app(lt(),app(s(),x)),app(s(),y)) -> app(app(lt(),x),y)
  app(app(lt(),|0|()),app(s(),y)) -> true()
  app(app(lt(),y),|0|()) -> false()
  app(app(eq(),x),x) -> true()
  app(app(eq(),app(s(),x)),|0|()) -> false()
  app(app(eq(),|0|()),app(s(),x)) -> false()
  app(app(member(),w),null()) -> false()
  app(app(member(),w),app(app(app(fork(),x),y),z)) -> app(app(app(if(),app(app(lt(),w),y)),app(app(member(),w),x)),app(app(app(if(),app(app(eq(),w),y)),true()),app(app(member(),w),z)))

Instead we prove termination of the uncurried system:

  app(eq_0(),y) -> eq_1(y)
  app(eq_1(x1),y) -> eq_2(x1,y)
  app(fork_0(),y) -> fork_1(y)
  app(fork_1(x1),y) -> fork_2(x1,y)
  app(fork_2(x1,x2),y) -> fork_3(x1,x2,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(lt_0(),y) -> lt_1(y)
  app(lt_1(x1),y) -> lt_2(x1,y)
  app(member_0(),y) -> member_1(y)
  app(member_1(x1),y) -> member_2(x1,y)
  app(s_0(),y) -> s_1(y)
  lt_2(s_1(x),s_1(y)) -> lt_2(x,y)
  lt_2(|0|_0(),s_1(y)) -> true_0()
  lt_2(y,|0|_0()) -> false_0()
  eq_2(x,x) -> true_0()
  eq_2(s_1(x),|0|_0()) -> false_0()
  eq_2(|0|_0(),s_1(x)) -> false_0()
  member_2(w,null_0()) -> false_0()
  member_2(w,fork_3(x,y,z)) -> if_3(lt_2(w,y),member_2(w,x),if_3(eq_2(w,y),true_0(),member_2(w,z)))


-- SCC decomposition.

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

p1: app#(eq_1(x1),y) -> eq_2#(x1,y)
p2: app#(lt_1(x1),y) -> lt_2#(x1,y)
p3: app#(member_1(x1),y) -> member_2#(x1,y)
p4: lt_2#(s_1(x),s_1(y)) -> lt_2#(x,y)
p5: member_2#(w,fork_3(x,y,z)) -> lt_2#(w,y)
p6: member_2#(w,fork_3(x,y,z)) -> member_2#(w,x)
p7: member_2#(w,fork_3(x,y,z)) -> eq_2#(w,y)
p8: member_2#(w,fork_3(x,y,z)) -> member_2#(w,z)

and R consists of:

r1: app(eq_0(),y) -> eq_1(y)
r2: app(eq_1(x1),y) -> eq_2(x1,y)
r3: app(fork_0(),y) -> fork_1(y)
r4: app(fork_1(x1),y) -> fork_2(x1,y)
r5: app(fork_2(x1,x2),y) -> fork_3(x1,x2,y)
r6: app(if_0(),y) -> if_1(y)
r7: app(if_1(x1),y) -> if_2(x1,y)
r8: app(if_2(x1,x2),y) -> if_3(x1,x2,y)
r9: app(lt_0(),y) -> lt_1(y)
r10: app(lt_1(x1),y) -> lt_2(x1,y)
r11: app(member_0(),y) -> member_1(y)
r12: app(member_1(x1),y) -> member_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: lt_2(s_1(x),s_1(y)) -> lt_2(x,y)
r15: lt_2(|0|_0(),s_1(y)) -> true_0()
r16: lt_2(y,|0|_0()) -> false_0()
r17: eq_2(x,x) -> true_0()
r18: eq_2(s_1(x),|0|_0()) -> false_0()
r19: eq_2(|0|_0(),s_1(x)) -> false_0()
r20: member_2(w,null_0()) -> false_0()
r21: member_2(w,fork_3(x,y,z)) -> if_3(lt_2(w,y),member_2(w,x),if_3(eq_2(w,y),true_0(),member_2(w,z)))

The estimated dependency graph contains the following SCCs:

  {p6, p8}
  {p4}


-- Reduction pair.

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

p1: member_2#(w,fork_3(x,y,z)) -> member_2#(w,z)
p2: member_2#(w,fork_3(x,y,z)) -> member_2#(w,x)

and R consists of:

r1: app(eq_0(),y) -> eq_1(y)
r2: app(eq_1(x1),y) -> eq_2(x1,y)
r3: app(fork_0(),y) -> fork_1(y)
r4: app(fork_1(x1),y) -> fork_2(x1,y)
r5: app(fork_2(x1,x2),y) -> fork_3(x1,x2,y)
r6: app(if_0(),y) -> if_1(y)
r7: app(if_1(x1),y) -> if_2(x1,y)
r8: app(if_2(x1,x2),y) -> if_3(x1,x2,y)
r9: app(lt_0(),y) -> lt_1(y)
r10: app(lt_1(x1),y) -> lt_2(x1,y)
r11: app(member_0(),y) -> member_1(y)
r12: app(member_1(x1),y) -> member_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: lt_2(s_1(x),s_1(y)) -> lt_2(x,y)
r15: lt_2(|0|_0(),s_1(y)) -> true_0()
r16: lt_2(y,|0|_0()) -> false_0()
r17: eq_2(x,x) -> true_0()
r18: eq_2(s_1(x),|0|_0()) -> false_0()
r19: eq_2(|0|_0(),s_1(x)) -> false_0()
r20: member_2(w,null_0()) -> false_0()
r21: member_2(w,fork_3(x,y,z)) -> if_3(lt_2(w,y),member_2(w,x),if_3(eq_2(w,y),true_0(),member_2(w,z)))

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:
      
        member_2#_A(x1,x2) = max{138, x1 + 1, x2 + 74}
        fork_3_A(x1,x2,x3) = max{x1 + 64, x2 - 73, x3}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            member_2#_A(x1,x2) = x2
            fork_3_A(x1,x2,x3) = x1 + x3 + 9
            member_2#!_A(x1,x2) = 9
            fork_3!_A(x1,x2,x3) = x1 + 9
  
    partial status:
  
      pi(member_2#) = [2]
      pi(fork_3) = [1, 3]

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: lt_2#(s_1(x),s_1(y)) -> lt_2#(x,y)

and R consists of:

r1: app(eq_0(),y) -> eq_1(y)
r2: app(eq_1(x1),y) -> eq_2(x1,y)
r3: app(fork_0(),y) -> fork_1(y)
r4: app(fork_1(x1),y) -> fork_2(x1,y)
r5: app(fork_2(x1,x2),y) -> fork_3(x1,x2,y)
r6: app(if_0(),y) -> if_1(y)
r7: app(if_1(x1),y) -> if_2(x1,y)
r8: app(if_2(x1,x2),y) -> if_3(x1,x2,y)
r9: app(lt_0(),y) -> lt_1(y)
r10: app(lt_1(x1),y) -> lt_2(x1,y)
r11: app(member_0(),y) -> member_1(y)
r12: app(member_1(x1),y) -> member_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: lt_2(s_1(x),s_1(y)) -> lt_2(x,y)
r15: lt_2(|0|_0(),s_1(y)) -> true_0()
r16: lt_2(y,|0|_0()) -> false_0()
r17: eq_2(x,x) -> true_0()
r18: eq_2(s_1(x),|0|_0()) -> false_0()
r19: eq_2(|0|_0(),s_1(x)) -> false_0()
r20: member_2(w,null_0()) -> false_0()
r21: member_2(w,fork_3(x,y,z)) -> if_3(lt_2(w,y),member_2(w,x),if_3(eq_2(w,y),true_0(),member_2(w,z)))

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:
      
        lt_2#_A(x1,x2) = 1
        s_1_A(x1) = 35
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            lt_2#_A(x1,x2) = x1 + x2
            s_1_A(x1) = x1 + 1
            lt_2#!_A(x1,x2) = x1
            s_1!_A(x1) = x1 + 8
  
    partial status:
  
      pi(lt_2#) = []
      pi(s_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.