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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        qsort#_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,0,0,0)) x1
        ._A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,1,1,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,0,1)) x2 + (2,0,1,0)
        lowers_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,0,0)) x2 + (1,2,3,1)
        greaters_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,1,0)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,1,1)) x2 + (1,1,6,0)
        nil_A() = (0,1,1,0)
        if_A(x1,x2,x3) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2 + ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,1,1,0)) x3 + (0,0,6,0)
        <=_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(1,1,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x2 + (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        . > qsort# > if > <= > greaters > lowers > nil
      
      argument filter:
    
        pi(qsort#) = []
        pi(.) = []
        pi(lowers) = []
        pi(greaters) = 2
        pi(nil) = []
        pi(if) = []
        pi(<=) = 2
    

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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        lowers#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,0,0),(1,0,0,0)) x2
        ._A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,1,0,0)) x2 + (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        . > lowers#
      
      argument filter:
    
        pi(lowers#) = [1]
        pi(.) = []
    

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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        greaters#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,0,0),(1,0,0,0)) x2
        ._A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,1,0,0)) x2 + (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        . > greaters#
      
      argument filter:
    
        pi(greaters#) = [1]
        pi(.) = []
    

The next rules are strictly ordered:

  p1

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