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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        nil > ifhigh > le > s > |0| > cons > false > low > iflow > high > true > quicksort#
      
      argument filter:
    
        pi(quicksort#) = 1
        pi(cons) = 2
        pi(high) = 2
        pi(low) = 2
        pi(le) = [1, 2]
        pi(|0|) = []
        pi(true) = []
        pi(s) = [1]
        pi(false) = []
        pi(iflow) = 3
        pi(ifhigh) = 3
        pi(nil) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        quicksort#_A(x1) = x1
        cons_A(x1,x2) = x2 + 2
        high_A(x1,x2) = x2 + 1
        low_A(x1,x2) = x2 + 1
        le_A(x1,x2) = 1
        |0|_A() = 0
        true_A() = 1
        s_A(x1) = x1 + 1
        false_A() = 1
        iflow_A(x1,x2,x3) = x3 + 1
        ifhigh_A(x1,x2,x3) = x3 + 1
        nil_A() = 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: 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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        cons > s > |0| > ifhigh# > high# > le > false > true
      
      argument filter:
    
        pi(ifhigh#) = 3
        pi(false) = []
        pi(cons) = [1, 2]
        pi(high#) = [2]
        pi(le) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        ifhigh#_A(x1,x2,x3) = 1
        false_A() = 1
        cons_A(x1,x2) = x1 + x2 + 2
        high#_A(x1,x2) = x2
        le_A(x1,x2) = 1
        true_A() = 1
        |0|_A() = 0
        s_A(x1) = x1 + 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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        cons > iflow# > s > le > |0| > true > low# > false
      
      argument filter:
    
        pi(iflow#) = 3
        pi(false) = []
        pi(cons) = [1, 2]
        pi(low#) = [2]
        pi(le) = 1
        pi(true) = []
        pi(|0|) = []
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        iflow#_A(x1,x2,x3) = 1
        false_A() = 1
        cons_A(x1,x2) = x1 + x2 + 2
        low#_A(x1,x2) = x2
        le_A(x1,x2) = 1
        true_A() = 1
        |0|_A() = 0
        s_A(x1) = x1 + 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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > le#
      
      argument filter:
    
        pi(le#) = 2
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        le#_A(x1,x2) = 0
        s_A(x1) = x1 + 1
    

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: 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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        cons > app#
      
      argument filter:
    
        pi(app#) = [1]
        pi(cons) = [1, 2]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1
        cons_A(x1,x2) = x1 + x2 + 1
    

The next rules are strictly ordered:

  p1

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