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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          qsort#_A(x1) = x1 + (1,1)
          ._A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(0,1)) x2 + (4,2)
          lowers_A(x1,x2) = ((1,1),(0,1)) x1 + (2,1)
          greaters_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(0,1)) x2 + (2,1)
          nil_A() = (1,1)
          if_A(x1,x2,x3) = (0,0)
          <=_A(x1,x2) = (3,2)
  
    precedence:
    
      nil > . = <= > lowers > greaters = if > qsort#
    
    partial status:
  
      pi(qsort#) = [1]
      pi(.) = [1]
      pi(lowers) = [1]
      pi(greaters) = []
      pi(nil) = []
      pi(if) = []
      pi(<=) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: qsort#(.(x,y)) -> qsort#(lowers(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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          qsort#_A(x1) = x1 + (1,0)
          ._A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(0,1)) x2 + (2,4)
          lowers_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(0,1)) x2 + (0,1)
          nil_A() = (1,2)
          if_A(x1,x2,x3) = x3 + (3,4)
          <=_A(x1,x2) = (3,5)
  
    precedence:
    
      qsort# = lowers = if > . = <= > nil
    
    partial status:
  
      pi(qsort#) = [1]
      pi(.) = [1]
      pi(lowers) = []
      pi(nil) = []
      pi(if) = []
      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: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          lowers#_A(x1,x2) = x2 + (0,1)
          ._A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
  
    precedence:
    
      . > lowers#
    
    partial status:
  
      pi(lowers#) = [2]
      pi(.) = [2]

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          greaters#_A(x1,x2) = x2 + (0,1)
          ._A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
  
    precedence:
    
      . > greaters#
    
    partial status:
  
      pi(greaters#) = [2]
      pi(.) = [2]

The next rules are strictly ordered:

  p1

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