YES

We show the termination of the TRS R:

  le(|0|(),Y) -> true()
  le(s(X),|0|()) -> false()
  le(s(X),s(Y)) -> le(X,Y)
  app(nil(),Y) -> Y
  app(cons(N,L),Y) -> cons(N,app(L,Y))
  low(N,nil()) -> nil()
  low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
  iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
  iflow(false(),N,cons(M,L)) -> low(N,L)
  high(N,nil()) -> nil()
  high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
  ifhigh(true(),N,cons(M,L)) -> high(N,L)
  ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
  quicksort(nil()) -> nil()
  quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

-- SCC decomposition.

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

p1: le#(s(X),s(Y)) -> le#(X,Y)
p2: app#(cons(N,L),Y) -> app#(L,Y)
p3: low#(N,cons(M,L)) -> iflow#(le(M,N),N,cons(M,L))
p4: low#(N,cons(M,L)) -> le#(M,N)
p5: iflow#(true(),N,cons(M,L)) -> low#(N,L)
p6: iflow#(false(),N,cons(M,L)) -> low#(N,L)
p7: high#(N,cons(M,L)) -> ifhigh#(le(M,N),N,cons(M,L))
p8: high#(N,cons(M,L)) -> le#(M,N)
p9: ifhigh#(true(),N,cons(M,L)) -> high#(N,L)
p10: ifhigh#(false(),N,cons(M,L)) -> high#(N,L)
p11: quicksort#(cons(N,L)) -> app#(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))
p12: quicksort#(cons(N,L)) -> quicksort#(low(N,L))
p13: quicksort#(cons(N,L)) -> low#(N,L)
p14: quicksort#(cons(N,L)) -> quicksort#(high(N,L))
p15: quicksort#(cons(N,L)) -> high#(N,L)

and R consists of:

r1: le(|0|(),Y) -> true()
r2: le(s(X),|0|()) -> false()
r3: le(s(X),s(Y)) -> le(X,Y)
r4: app(nil(),Y) -> Y
r5: app(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The estimated dependency graph contains the following SCCs:

  {p12, p14}
  {p7, p9, p10}
  {p3, p5, p6}
  {p1}
  {p2}


-- Reduction pair.

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

p1: quicksort#(cons(N,L)) -> quicksort#(high(N,L))
p2: quicksort#(cons(N,L)) -> quicksort#(low(N,L))

and R consists of:

r1: le(|0|(),Y) -> true()
r2: le(s(X),|0|()) -> false()
r3: le(s(X),s(Y)) -> le(X,Y)
r4: app(nil(),Y) -> Y
r5: app(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8, r9, r10, r11, r12, r13

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      quicksort#_A(x1) = x1
      cons_A(x1,x2) = x2 + (2,1)
      high_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (1,3)
      low_A(x1,x2) = x2 + (1,3)
      le_A(x1,x2) = (2,1)
      |0|_A() = (1,1)
      true_A() = (1,2)
      s_A(x1) = ((0,0),(1,0)) x1 + (1,1)
      false_A() = (1,2)
      iflow_A(x1,x2,x3) = x3 + (1,2)
      ifhigh_A(x1,x2,x3) = ((0,0),(1,0)) x2 + x3 + (1,2)
      nil_A() = (1,4)

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: ifhigh#(false(),N,cons(M,L)) -> high#(N,L)
p2: high#(N,cons(M,L)) -> ifhigh#(le(M,N),N,cons(M,L))
p3: ifhigh#(true(),N,cons(M,L)) -> high#(N,L)

and R consists of:

r1: le(|0|(),Y) -> true()
r2: le(s(X),|0|()) -> false()
r3: le(s(X),s(Y)) -> le(X,Y)
r4: app(nil(),Y) -> Y
r5: app(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      ifhigh#_A(x1,x2,x3) = ((0,0),(1,0)) x2 + x3 + (0,1)
      false_A() = (1,1)
      cons_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,1)
      high#_A(x1,x2) = x2 + (1,0)
      le_A(x1,x2) = ((1,0),(1,1)) x2 + (2,1)
      true_A() = (1,2)
      |0|_A() = (1,1)
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1, p2, p3

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: iflow#(false(),N,cons(M,L)) -> low#(N,L)
p2: low#(N,cons(M,L)) -> iflow#(le(M,N),N,cons(M,L))
p3: iflow#(true(),N,cons(M,L)) -> low#(N,L)

and R consists of:

r1: le(|0|(),Y) -> true()
r2: le(s(X),|0|()) -> false()
r3: le(s(X),s(Y)) -> le(X,Y)
r4: app(nil(),Y) -> Y
r5: app(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      iflow#_A(x1,x2,x3) = ((0,0),(1,0)) x2 + x3
      false_A() = (1,5)
      cons_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,1)
      low#_A(x1,x2) = ((1,0),(1,1)) x2 + (1,2)
      le_A(x1,x2) = ((1,0),(1,1)) x2 + (2,2)
      true_A() = (1,1)
      |0|_A() = (1,1)
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1, p2, p3

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: le#(s(X),s(Y)) -> le#(X,Y)

and R consists of:

r1: le(|0|(),Y) -> true()
r2: le(s(X),|0|()) -> false()
r3: le(s(X),s(Y)) -> le(X,Y)
r4: app(nil(),Y) -> Y
r5: app(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      le#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

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#(cons(N,L),Y) -> app#(L,Y)

and R consists of:

r1: le(|0|(),Y) -> true()
r2: le(s(X),|0|()) -> false()
r3: le(s(X),s(Y)) -> le(X,Y)
r4: app(nil(),Y) -> Y
r5: app(cons(N,L),Y) -> cons(N,app(L,Y))
r6: low(N,nil()) -> nil()
r7: low(N,cons(M,L)) -> iflow(le(M,N),N,cons(M,L))
r8: iflow(true(),N,cons(M,L)) -> cons(M,low(N,L))
r9: iflow(false(),N,cons(M,L)) -> low(N,L)
r10: high(N,nil()) -> nil()
r11: high(N,cons(M,L)) -> ifhigh(le(M,N),N,cons(M,L))
r12: ifhigh(true(),N,cons(M,L)) -> high(N,L)
r13: ifhigh(false(),N,cons(M,L)) -> cons(M,high(N,L))
r14: quicksort(nil()) -> nil()
r15: quicksort(cons(N,L)) -> app(quicksort(low(N,L)),cons(N,quicksort(high(N,L))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0),(1,0)) x1
      cons_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)

The next rules are strictly ordered:

  p1

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