YES

We show the termination of the TRS R:

  qsort(nil()) -> nil()
  qsort(.(x,y)) -> ++(qsort(lowers(x,y)),.(x,qsort(greaters(x,y))))
  lowers(x,nil()) -> nil()
  lowers(x,.(y,z)) -> if(<=(y,x),.(y,lowers(x,z)),lowers(x,z))
  greaters(x,nil()) -> nil()
  greaters(x,.(y,z)) -> if(<=(y,x),greaters(x,z),.(y,greaters(x,z)))

-- SCC decomposition.

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

p1: qsort#(.(x,y)) -> qsort#(lowers(x,y))
p2: qsort#(.(x,y)) -> lowers#(x,y)
p3: qsort#(.(x,y)) -> qsort#(greaters(x,y))
p4: qsort#(.(x,y)) -> greaters#(x,y)
p5: lowers#(x,.(y,z)) -> lowers#(x,z)
p6: greaters#(x,.(y,z)) -> greaters#(x,z)

and R consists of:

r1: qsort(nil()) -> nil()
r2: qsort(.(x,y)) -> ++(qsort(lowers(x,y)),.(x,qsort(greaters(x,y))))
r3: lowers(x,nil()) -> nil()
r4: lowers(x,.(y,z)) -> if(<=(y,x),.(y,lowers(x,z)),lowers(x,z))
r5: greaters(x,nil()) -> nil()
r6: greaters(x,.(y,z)) -> if(<=(y,x),greaters(x,z),.(y,greaters(x,z)))

The estimated dependency graph contains the following SCCs:

  {p1, p3}
  {p5}
  {p6}


-- Reduction pair.

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

p1: qsort#(.(x,y)) -> qsort#(lowers(x,y))
p2: qsort#(.(x,y)) -> qsort#(greaters(x,y))

and R consists of:

r1: qsort(nil()) -> nil()
r2: qsort(.(x,y)) -> ++(qsort(lowers(x,y)),.(x,qsort(greaters(x,y))))
r3: lowers(x,nil()) -> nil()
r4: lowers(x,.(y,z)) -> if(<=(y,x),.(y,lowers(x,z)),lowers(x,z))
r5: greaters(x,nil()) -> nil()
r6: greaters(x,.(y,z)) -> if(<=(y,x),greaters(x,z),.(y,greaters(x,z)))

The set of usable rules consists of

  r3, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      qsort#_A(x1) = x1
      ._A(x1,x2) = max{x1 + 9, x2 + 4}
      lowers_A(x1,x2) = max{x1 + 2, x2 - 7}
      greaters_A(x1,x2) = max{x1 + 5, x2 - 1}
      nil_A = 0
      if_A(x1,x2,x3) = max{2, x1 - 4, x2 - 13, x3 - 5}
      <=_A(x1,x2) = max{x1 + 6, x2 + 6}
    
    2. max/plus interpretations on natural numbers:
    
      qsort#_A(x1) = max{0, x1 - 4}
      ._A(x1,x2) = 5
      lowers_A(x1,x2) = 3
      greaters_A(x1,x2) = 1
      nil_A = 0
      if_A(x1,x2,x3) = 2
      <=_A(x1,x2) = 0
    

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: lowers#(x,.(y,z)) -> lowers#(x,z)

and R consists of:

r1: qsort(nil()) -> nil()
r2: qsort(.(x,y)) -> ++(qsort(lowers(x,y)),.(x,qsort(greaters(x,y))))
r3: lowers(x,nil()) -> nil()
r4: lowers(x,.(y,z)) -> if(<=(y,x),.(y,lowers(x,z)),lowers(x,z))
r5: greaters(x,nil()) -> nil()
r6: greaters(x,.(y,z)) -> if(<=(y,x),greaters(x,z),.(y,greaters(x,z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      lowers#_A(x1,x2) = x2
      ._A(x1,x2) = max{x1 + 1, x2 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      lowers#_A(x1,x2) = 0
      ._A(x1,x2) = 0
    

The next rules are strictly ordered:

  p1

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: greaters#(x,.(y,z)) -> greaters#(x,z)

and R consists of:

r1: qsort(nil()) -> nil()
r2: qsort(.(x,y)) -> ++(qsort(lowers(x,y)),.(x,qsort(greaters(x,y))))
r3: lowers(x,nil()) -> nil()
r4: lowers(x,.(y,z)) -> if(<=(y,x),.(y,lowers(x,z)),lowers(x,z))
r5: greaters(x,nil()) -> nil()
r6: greaters(x,.(y,z)) -> if(<=(y,x),greaters(x,z),.(y,greaters(x,z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      greaters#_A(x1,x2) = x2
      ._A(x1,x2) = max{x1 + 1, x2 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      greaters#_A(x1,x2) = 0
      ._A(x1,x2) = 0
    

The next rules are strictly ordered:

  p1

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