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, r15, r16, r17, r19

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 = 47
            quicksort#_A(x1) = 1
            isempty_A(x1) = x1
            low_A(x1,x2) = x2 + 6
            head_A(x1) = x1 + 11
            tail_A(x1) = max{0, x1 - 14}
            high_A(x1,x2) = max{17, x2 + 5}
            le_A(x1,x2) = 47
            |0|_A = 34
            true_A = 25
            s_A(x1) = max{34, x1 + 33}
            nil_A = 42
            add_A(x1,x2) = max{96, x1 + 46, x2 + 44}
            if_low_A(x1,x2,x3) = max{96, x1 - 24, x3 + 6}
            if_high_A(x1,x2,x3) = max{96, x1 - 48, x3 + 5}
            if_qs_A(x1,x2,x3,x4) = max{42, x1 - 53, x2 + 3, x3 - 8, x4 - 18}
            if_qs#!_A(x1,x2,x3,x4) = max{x1 - 8, x2 - 7, x4 + 1}
            false!_A = 31
            quicksort#!_A(x1) = max{29, x1 - 7}
            isempty!_A(x1) = max{30, x1 - 7}
            low!_A(x1,x2) = 26
            head!_A(x1) = max{32, x1 - 8}
            tail!_A(x1) = max{32, x1 - 8}
            high!_A(x1,x2) = max{28, x2 - 43}
            le!_A(x1,x2) = max{21, x1 - 1, x2 + 1}
            |0|!_A = 35
            true!_A = 34
            s!_A(x1) = max{32, x1 - 2}
            nil!_A = 27
            add!_A(x1,x2) = 22
            if_low!_A(x1,x2,x3) = max{26, x1 - 22}
            if_high!_A(x1,x2,x3) = max{29, x1 - 24, x3 - 69}
            if_qs!_A(x1,x2,x3,x4) = max{32, x1 + 1, x3 + 28}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_qs#_A(x1,x2,x3,x4) = 1
            false_A = 27
            quicksort#_A(x1) = 1
            isempty_A(x1) = max{27, x1 + 7}
            low_A(x1,x2) = 24
            head_A(x1) = x1 + 7
            tail_A(x1) = max{72, x1 + 11}
            high_A(x1,x2) = 0
            le_A(x1,x2) = max{4, x1 - 26, x2 - 1}
            |0|_A = 27
            true_A = 2
            s_A(x1) = max{53, x1 + 27}
            nil_A = 0
            add_A(x1,x2) = max{0, x2 - 11}
            if_low_A(x1,x2,x3) = 24
            if_high_A(x1,x2,x3) = 0
            if_qs_A(x1,x2,x3,x4) = max{0, x1 - 64, x2 - 84, x3 - 10, x4 - 37}
            if_qs#!_A(x1,x2,x3,x4) = max{0, x1 - 26, x2 - 24, x4 - 1}
            false!_A = 9
            quicksort#!_A(x1) = max{0, x1 - 25}
            isempty!_A(x1) = max{21, x1 - 27}
            low!_A(x1,x2) = 8
            head!_A(x1) = max{1, x1 - 24}
            tail!_A(x1) = max{23, x1 - 25}
            high!_A(x1,x2) = max{48, x2 + 46}
            le!_A(x1,x2) = max{x1 - 8, x2 + 19}
            |0|!_A = 21
            true!_A = 20
            s!_A(x1) = max{10, x1 - 9}
            nil!_A = 7
            add!_A(x1,x2) = 0
            if_low!_A(x1,x2,x3) = 8
            if_high!_A(x1,x2,x3) = max{47, x1 - 28, x2 - 1}
            if_qs!_A(x1,x2,x3,x4) = max{x1 - 6, x2 + 23, x3 + 1, x4 + 6}
        
  
    partial status:
  
      pi(if_qs#) = [4]
      pi(false) = []
      pi(quicksort#) = []
      pi(isempty) = [1]
      pi(low) = [2]
      pi(head) = []
      pi(tail) = [1]
      pi(high) = []
      pi(le) = [1, 2]
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]
      pi(nil) = []
      pi(add) = [2]
      pi(if_low) = [3]
      pi(if_high) = []
      pi(if_qs) = [1, 2, 3, 4]

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_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 - 24, x2 - 3, x3 - 57}
            false_A = 1
            add_A(x1,x2) = max{55, x1 + 20, x2 + 20}
            high#_A(x1,x2) = max{0, x1 - 3, x2 - 57}
            le_A(x1,x2) = max{23, x1 - 13}
            true_A = 23
            |0|_A = 2
            s_A(x1) = x1 + 12
            if_high#!_A(x1,x2,x3) = max{6, x1 - 47, x3 - 52}
            false!_A = 3
            add!_A(x1,x2) = max{4, x1 - 432, x2 - 25}
            high#!_A(x1,x2) = max{5, x2 - 46}
            le!_A(x1,x2) = max{8, x1 - 3, x2 - 13}
            true!_A = 1
            |0|!_A = 4
            s!_A(x1) = max{4, x1 - 14}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_high#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 16, x3 - 37}
            false_A = 1
            add_A(x1,x2) = max{x1 + 35, x2 + 10}
            high#_A(x1,x2) = max{0, x1 - 16, x2 - 37}
            le_A(x1,x2) = max{1, x1 - 1, x2 - 26}
            true_A = 0
            |0|_A = 0
            s_A(x1) = x1
            if_high#!_A(x1,x2,x3) = max{5, x1 - 39, x3 - 83}
            false!_A = 1
            add!_A(x1,x2) = max{4, x1 - 47, x2 - 39}
            high#!_A(x1,x2) = max{x1 + 8, x2 - 74}
            le!_A(x1,x2) = max{7, x1 - 1, x2 - 3}
            true!_A = 6
            |0|!_A = 7
            s!_A(x1) = max{2, x1 - 2}
        
  
    partial status:
  
      pi(if_high#) = [2]
      pi(false) = []
      pi(add) = [1]
      pi(high#) = [1]
      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, 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_low#_A(x1,x2,x3) = max{0, x1 - 18, x2 - 19, x3 - 23}
            false_A = 17
            add_A(x1,x2) = max{x1 + 4, x2 + 14}
            low#_A(x1,x2) = max{0, x1 - 19, x2 - 16}
            le_A(x1,x2) = max{17, x1 - 3}
            true_A = 11
            |0|_A = 2
            s_A(x1) = x1
            isempty_A(x1) = x1 + 12
            if_low#!_A(x1,x2,x3) = max{0, x1 - 16, x3 - 15}
            false!_A = 28
            add!_A(x1,x2) = max{36, x1 + 2, x2 + 26}
            low#!_A(x1,x2) = max{0, x2 - 2}
            le!_A(x1,x2) = max{37, x1 - 3, x2 - 128}
            true!_A = 1
            |0|!_A = 29
            s!_A(x1) = max{29, x1 - 127}
            isempty!_A(x1) = max{30, x1 - 26}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_low#_A(x1,x2,x3) = max{0, x1 - 62, x2 - 67, x3 - 13}
            false_A = 2
            add_A(x1,x2) = max{12, x1 - 54, x2 + 5}
            low#_A(x1,x2) = max{0, x1 - 67, x2 - 13}
            le_A(x1,x2) = max{11, x1 - 5, x2 - 5}
            true_A = 0
            |0|_A = 4
            s_A(x1) = x1
            isempty_A(x1) = max{38, x1 - 17}
            if_low#!_A(x1,x2,x3) = max{x1 - 1, x2, x3 - 14}
            false!_A = 6
            add!_A(x1,x2) = max{9, x1 - 69, x2 - 69}
            low#!_A(x1,x2) = max{0, x2 - 74}
            le!_A(x1,x2) = max{10, x1 - 1, x2 - 5}
            true!_A = 2
            |0|!_A = 7
            s!_A(x1) = max{7, x1 - 6}
            isempty!_A(x1) = max{7, x1 - 17}
        
  
    partial status:
  
      pi(if_low#) = []
      pi(false) = []
      pi(add) = []
      pi(low#) = []
      pi(le) = [1]
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(isempty) = [1]

The following rules are removed:

  p1, p2



-- SCC decomposition.

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

p1: 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 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:

  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) = x1 + 29
            le#!_A(x1,x2) = max{6, x1 - 30, x2 - 206}
            s!_A(x1) = max{5, x1 - 206}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = max{x1, x2}
            s_A(x1) = x1 + 23
            le#!_A(x1,x2) = max{6, x1 - 64, x2 - 66}
            s!_A(x1) = max{5, x1 - 65}
        
  
    partial status:
  
      pi(le#) = [2]
      pi(s) = [1]

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

  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:
          
            app#_A(x1,x2) = 0
            add_A(x1,x2) = max{x1 + 6, x2 - 4}
            isempty_A(x1) = max{23, x1 - 16}
            false_A = 1
            app#!_A(x1,x2) = 0
            add!_A(x1,x2) = max{3, x1 - 57}
            isempty!_A(x1) = max{8, x1 - 26}
            false!_A = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 0
            add_A(x1,x2) = max{6, x1 + 4, x2 - 9}
            isempty_A(x1) = max{19, x1 - 10}
            false_A = 1
            app#!_A(x1,x2) = max{1, x1 - 81, x2 - 1}
            add!_A(x1,x2) = max{4, x1 - 77, x2 - 81}
            isempty!_A(x1) = max{6, x1 - 10}
            false!_A = 5
        
  
    partial status:
  
      pi(app#) = [1]
      pi(add) = [1, 2]
      pi(isempty) = [1]
      pi(false) = []

The following rules are removed:

  p1

Then no dependency pair remains.