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)))

-- SCC decomposition.

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

p1: app#(app(lt(),app(s(),x)),app(s(),y)) -> app#(app(lt(),x),y)
p2: app#(app(lt(),app(s(),x)),app(s(),y)) -> app#(lt(),x)
p3: 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)))
p4: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(app(if(),app(app(lt(),w),y)),app(app(member(),w),x))
p5: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(if(),app(app(lt(),w),y))
p6: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(app(lt(),w),y)
p7: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(lt(),w)
p8: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(app(member(),w),x)
p9: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(app(app(if(),app(app(eq(),w),y)),true()),app(app(member(),w),z))
p10: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(app(if(),app(app(eq(),w),y)),true())
p11: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(if(),app(app(eq(),w),y))
p12: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(app(eq(),w),y)
p13: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(eq(),w)
p14: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(app(member(),w),z)

and R consists of:

r1: app(app(lt(),app(s(),x)),app(s(),y)) -> app(app(lt(),x),y)
r2: app(app(lt(),|0|()),app(s(),y)) -> true()
r3: app(app(lt(),y),|0|()) -> false()
r4: app(app(eq(),x),x) -> true()
r5: app(app(eq(),app(s(),x)),|0|()) -> false()
r6: app(app(eq(),|0|()),app(s(),x)) -> false()
r7: app(app(member(),w),null()) -> false()
r8: 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)))

The estimated dependency graph contains the following SCCs:

  {p8, p14}
  {p1}


-- Reduction pair.

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

p1: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(app(member(),w),z)
p2: app#(app(member(),w),app(app(app(fork(),x),y),z)) -> app#(app(member(),w),x)

and R consists of:

r1: app(app(lt(),app(s(),x)),app(s(),y)) -> app(app(lt(),x),y)
r2: app(app(lt(),|0|()),app(s(),y)) -> true()
r3: app(app(lt(),y),|0|()) -> false()
r4: app(app(eq(),x),x) -> true()
r5: app(app(eq(),app(s(),x)),|0|()) -> false()
r6: app(app(eq(),|0|()),app(s(),x)) -> false()
r7: app(app(member(),w),null()) -> false()
r8: 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)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1 + ((1,0),(1,1)) x2
        app_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,1),(1,0)) x2
        member_A() = (0,0)
        fork_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1 + ((0,1),(1,0)) x2
        app_A(x1,x2) = ((1,1),(0,0)) x1 + ((0,1),(1,1)) x2 + (0,1)
        member_A() = (0,0)
        fork_A() = (1,1)
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

p1: app#(app(lt(),app(s(),x)),app(s(),y)) -> app#(app(lt(),x),y)

and R consists of:

r1: app(app(lt(),app(s(),x)),app(s(),y)) -> app(app(lt(),x),y)
r2: app(app(lt(),|0|()),app(s(),y)) -> true()
r3: app(app(lt(),y),|0|()) -> false()
r4: app(app(eq(),x),x) -> true()
r5: app(app(eq(),app(s(),x)),|0|()) -> false()
r6: app(app(eq(),|0|()),app(s(),x)) -> false()
r7: app(app(member(),w),null()) -> false()
r8: 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)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        app#_A(x1,x2) = ((1,1),(0,0)) x1
        app_A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (1,0)
        lt_A() = (1,0)
        s_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        app#_A(x1,x2) = (0,0)
        app_A(x1,x2) = (1,1)
        lt_A() = (1,1)
        s_A() = (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.