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(add(n,x),y) -> add(n,app(x,y))
  low(n,nil()) -> nil()
  low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
  if_low(true(),n,add(m,x)) -> add(m,low(n,x))
  if_low(false(),n,add(m,x)) -> low(n,x)
  high(n,nil()) -> nil()
  high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
  if_high(true(),n,add(m,x)) -> high(n,x)
  if_high(false(),n,add(m,x)) -> add(m,high(n,x))
  head(add(n,x)) -> n
  tail(add(n,x)) -> x
  isempty(nil()) -> true()
  isempty(add(n,x)) -> false()
  quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
  if_qs(true(),x,n,y) -> nil()
  if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

-- SCC decomposition.

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

p1: le#(s(x),s(y)) -> le#(x,y)
p2: app#(add(n,x),y) -> app#(x,y)
p3: low#(n,add(m,x)) -> if_low#(le(m,n),n,add(m,x))
p4: low#(n,add(m,x)) -> le#(m,n)
p5: if_low#(true(),n,add(m,x)) -> low#(n,x)
p6: if_low#(false(),n,add(m,x)) -> low#(n,x)
p7: high#(n,add(m,x)) -> if_high#(le(m,n),n,add(m,x))
p8: high#(n,add(m,x)) -> le#(m,n)
p9: if_high#(true(),n,add(m,x)) -> high#(n,x)
p10: if_high#(false(),n,add(m,x)) -> high#(n,x)
p11: quicksort#(x) -> if_qs#(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
p12: quicksort#(x) -> isempty#(x)
p13: quicksort#(x) -> low#(head(x),tail(x))
p14: quicksort#(x) -> head#(x)
p15: quicksort#(x) -> tail#(x)
p16: quicksort#(x) -> high#(head(x),tail(x))
p17: if_qs#(false(),x,n,y) -> app#(quicksort(x),add(n,quicksort(y)))
p18: if_qs#(false(),x,n,y) -> quicksort#(x)
p19: if_qs#(false(),x,n,y) -> quicksort#(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(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The estimated dependency graph contains the following SCCs:

  {p11, p18, p19}
  {p7, p9, p10}
  {p3, p5, p6}
  {p1}
  {p2}


-- Reduction pair.

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

p1: if_qs#(false(),x,n,y) -> quicksort#(y)
p2: quicksort#(x) -> if_qs#(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
p3: if_qs#(false(),x,n,y) -> quicksort#(x)

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(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    if_qs#_A(x1,x2,x3,x4) = max{x1 - 1, x2 + 1, x3 - 2, x4 + 2}
    false_A = 7
    quicksort#_A(x1) = max{5, x1 + 1}
    isempty_A(x1) = max{1, x1 - 3}
    low_A(x1,x2) = max{3, x2}
    head_A(x1) = x1 + 2
    tail_A(x1) = max{0, x1 - 10}
    high_A(x1,x2) = max{2, x2}
    le_A(x1,x2) = max{0, x1 - 2}
    |0|_A = 3
    true_A = 1
    s_A(x1) = x1 + 9
    if_low_A(x1,x2,x3) = max{x1 + 12, x3}
    add_A(x1,x2) = max{12, x1 + 10, x2 + 10}
    if_high_A(x1,x2,x3) = max{11, x3}
    nil_A = 3

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.

-- SCC decomposition.

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

p1: if_qs#(false(),x,n,y) -> quicksort#(x)

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(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: if_high#(false(),n,add(m,x)) -> high#(n,x)
p2: high#(n,add(m,x)) -> if_high#(le(m,n),n,add(m,x))
p3: if_high#(true(),n,add(m,x)) -> high#(n,x)

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(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    if_high#_A(x1,x2,x3) = max{x1 + 2, x3 - 3}
    false_A = 1
    add_A(x1,x2) = max{x1 + 5, x2 + 4}
    high#_A(x1,x2) = max{2, x2}
    le_A(x1,x2) = 1
    true_A = 0
    |0|_A = 0
    s_A(x1) = max{0, x1 - 1}

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.

-- SCC decomposition.

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

p1: if_high#(true(),n,add(m,x)) -> high#(n,x)

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(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: if_low#(false(),n,add(m,x)) -> low#(n,x)
p2: low#(n,add(m,x)) -> if_low#(le(m,n),n,add(m,x))
p3: if_low#(true(),n,add(m,x)) -> low#(n,x)

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(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    if_low#_A(x1,x2,x3) = max{1, x1 - 1, x3}
    false_A = 0
    add_A(x1,x2) = max{x1 + 1, x2 + 1}
    low#_A(x1,x2) = x2
    le_A(x1,x2) = 2
    true_A = 2
    |0|_A = 0
    s_A(x1) = max{0, x1 - 1}

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: low#(n,add(m,x)) -> if_low#(le(m,n),n,add(m,x))

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(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    le#_A(x1,x2) = max{x1 + 1, x2 + 1}
    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#(add(n,x),y) -> app#(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(add(n,x),y) -> add(n,app(x,y))
r6: low(n,nil()) -> nil()
r7: low(n,add(m,x)) -> if_low(le(m,n),n,add(m,x))
r8: if_low(true(),n,add(m,x)) -> add(m,low(n,x))
r9: if_low(false(),n,add(m,x)) -> low(n,x)
r10: high(n,nil()) -> nil()
r11: high(n,add(m,x)) -> if_high(le(m,n),n,add(m,x))
r12: if_high(true(),n,add(m,x)) -> high(n,x)
r13: if_high(false(),n,add(m,x)) -> add(m,high(n,x))
r14: head(add(n,x)) -> n
r15: tail(add(n,x)) -> x
r16: isempty(nil()) -> true()
r17: isempty(add(n,x)) -> false()
r18: quicksort(x) -> if_qs(isempty(x),low(head(x),tail(x)),head(x),high(head(x),tail(x)))
r19: if_qs(true(),x,n,y) -> nil()
r20: if_qs(false(),x,n,y) -> app(quicksort(x),add(n,quicksort(y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    app#_A(x1,x2) = x1
    add_A(x1,x2) = max{x1 + 1, x2 + 1}

The next rules are strictly ordered:

  p1

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