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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_qs#_A(x1,x2,x3,x4) = 1
            false_A = 25
            quicksort#_A(x1) = 1
            isempty_A(x1) = max{0, x1 - 9}
            low_A(x1,x2) = max{3, x2 + 1}
            head_A(x1) = max{18, x1 + 1}
            tail_A(x1) = max{6, x1 - 10}
            high_A(x1,x2) = max{2, x2 + 1}
            le_A(x1,x2) = max{25, x1 - 8, x2 - 9}
            |0|_A = 5
            true_A = 0
            s_A(x1) = x1 + 1
            nil_A = 8
            add_A(x1,x2) = max{x1 + 34, x2 + 32}
            if_low_A(x1,x2,x3) = max{34, x3 + 1}
            if_high_A(x1,x2,x3) = x3 + 1
            if_qs#!_A(x1,x2,x3,x4) = max{x1 - 1, x2 + 16, x3 + 1, x4 + 9}
            false!_A = 2
            quicksort#!_A(x1) = max{24, x1 + 8}
            isempty!_A(x1) = max{7, x1 - 2}
            low!_A(x1,x2) = max{25, x1}
            head!_A(x1) = max{25, x1 - 1}
            tail!_A(x1) = 72
            high!_A(x1,x2) = max{x1 + 53, x2 + 19}
            le!_A(x1,x2) = max{x1 + 54, x2 - 2}
            |0|!_A = 6
            true!_A = 7
            s!_A(x1) = max{53, x1 - 3}
            nil!_A = 26
            add!_A(x1,x2) = 25
            if_low!_A(x1,x2,x3) = max{25, x2}
            if_high!_A(x1,x2,x3) = max{x1 + 8, x2 + 53, x3 - 6}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if_qs#_A(x1,x2,x3,x4) = 0
              false_A() = 1
              quicksort#_A(x1) = 0
              isempty_A(x1) = 1
              low_A(x1,x2) = x2
              head_A(x1) = x1 + 1
              tail_A(x1) = x1 + 1
              high_A(x1,x2) = x1 + x2
              le_A(x1,x2) = 1
              |0|_A() = 1
              true_A() = 1
              s_A(x1) = x1 + 4
              nil_A() = 2
              add_A(x1,x2) = x1 + x2 + 2
              if_low_A(x1,x2,x3) = x3
              if_high_A(x1,x2,x3) = x2 + x3
              if_qs#!_A(x1,x2,x3,x4) = 6
              false!_A() = 4
              quicksort#!_A(x1) = 6
              isempty!_A(x1) = 5
              low!_A(x1,x2) = x2 + 6
              head!_A(x1) = 4
              tail!_A(x1) = 3
              high!_A(x1,x2) = x2 + 3
              le!_A(x1,x2) = x1 + x2
              |0|!_A() = 1
              true!_A() = 2
              s!_A(x1) = 5
              nil!_A() = 0
              add!_A(x1,x2) = x1 + x2 + 5
              if_low!_A(x1,x2,x3) = x3 + 5
              if_high!_A(x1,x2,x3) = x1 + x3 + 1
        
  
    partial status:
  
      pi(if_qs#) = []
      pi(false) = []
      pi(quicksort#) = []
      pi(isempty) = [1]
      pi(low) = [2]
      pi(head) = [1]
      pi(tail) = [1]
      pi(high) = [2]
      pi(le) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]
      pi(nil) = []
      pi(add) = [1, 2]
      pi(if_low) = [3]
      pi(if_high) = [3]

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: if_qs#(false(),x,n,y) -> quicksort#(y)
p2: 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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_high#_A(x1,x2,x3) = max{0, x1 - 14, x2 - 3, x3 - 18}
            false_A = 1
            add_A(x1,x2) = max{16, x1 + 15, x2 + 7}
            high#_A(x1,x2) = max{0, x1 - 3, x2 - 18}
            le_A(x1,x2) = max{13, x1 - 9, x2 - 14}
            true_A = 1
            |0|_A = 8
            s_A(x1) = max{2, x1 + 1}
            if_high#!_A(x1,x2,x3) = max{14, x1 - 67, x2 + 5, x3 - 17}
            false!_A = 2
            add!_A(x1,x2) = max{4, x1 - 53, x2 - 73}
            high#!_A(x1,x2) = max{14, x1 + 5, x2 - 10}
            le!_A(x1,x2) = max{13, x1 + 5}
            true!_A = 12
            |0|!_A = 13
            s!_A(x1) = x1 + 14
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if_high#_A(x1,x2,x3) = x3
              false_A() = 2
              add_A(x1,x2) = x1 + x2 + 5
              high#_A(x1,x2) = x2 + 5
              le_A(x1,x2) = x1 + x2 + 1
              true_A() = 2
              |0|_A() = 1
              s_A(x1) = x1 + 1
              if_high#!_A(x1,x2,x3) = x3
              false!_A() = 0
              add!_A(x1,x2) = x1 + 6
              high#!_A(x1,x2) = x2 + 2
              le!_A(x1,x2) = x1 + x2 + 8
              true!_A() = 5
              |0|!_A() = 6
              s!_A(x1) = 1
        
  
    partial status:
  
      pi(if_high#) = []
      pi(false) = []
      pi(add) = []
      pi(high#) = []
      pi(le) = [1]
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- 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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_low#_A(x1,x2,x3) = max{0, x1 - 18, x2 - 11, x3 - 41}
            false_A = 4
            add_A(x1,x2) = max{x1 + 37, x2 + 39}
            low#_A(x1,x2) = max{0, x1 - 11, x2 - 41}
            le_A(x1,x2) = 17
            true_A = 4
            |0|_A = 0
            s_A(x1) = 0
            if_low#!_A(x1,x2,x3) = max{0, x1 - 5}
            false!_A = 3
            add!_A(x1,x2) = max{x1 + 4, x2 - 17}
            low#!_A(x1,x2) = max{4, x2 - 18}
            le!_A(x1,x2) = 11
            true!_A = 10
            |0|!_A = 11
            s!_A(x1) = max{12, x1 - 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if_low#_A(x1,x2,x3) = x3
              false_A() = 2
              add_A(x1,x2) = x1 + x2 + 2
              low#_A(x1,x2) = x2 + 2
              le_A(x1,x2) = x1 + x2 + 1
              true_A() = 2
              |0|_A() = 4
              s_A(x1) = x1 + 1
              if_low#!_A(x1,x2,x3) = x3 + 2
              false!_A() = 4
              add!_A(x1,x2) = 3
              low#!_A(x1,x2) = 3
              le!_A(x1,x2) = x1 + 7
              true!_A() = 10
              |0|!_A() = 11
              s!_A(x1) = 5
        
  
    partial status:
  
      pi(if_low#) = [3]
      pi(false) = []
      pi(add) = []
      pi(low#) = []
      pi(le) = [2]
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p1, p2, p3

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(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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = max{x1, x2}
            s_A(x1) = max{2, x1 + 1}
            le#!_A(x1,x2) = 1
            s!_A(x1) = max{6, x1 - 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              le#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              le#!_A(x1,x2) = x2
              s!_A(x1) = x1 + 3
        
  
    partial status:
  
      pi(le#) = []
      pi(s) = []

The following rules are removed:

  p1

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 0
            add_A(x1,x2) = max{x1 + 20, x2 + 12}
            app#!_A(x1,x2) = max{1, x1 - 10}
            add!_A(x1,x2) = max{7, x1 + 5, x2 + 5}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = x1
              add_A(x1,x2) = x2 + 2
              app#!_A(x1,x2) = 3
              add!_A(x1,x2) = 4
        
  
    partial status:
  
      pi(app#) = []
      pi(add) = []

The following rules are removed:

  p1

Then no dependency pair remains.