YES

We show the termination of the TRS R:

  lt(|0|(),s(X)) -> true()
  lt(s(X),|0|()) -> false()
  lt(s(X),s(Y)) -> lt(X,Y)
  append(nil(),Y) -> Y
  append(add(N,X),Y) -> add(N,append(X,Y))
  split(N,nil()) -> pair(nil(),nil())
  split(N,add(M,Y)) -> f_1(split(N,Y),N,M,Y)
  f_1(pair(X,Z),N,M,Y) -> f_2(lt(N,M),N,M,Y,X,Z)
  f_2(true(),N,M,Y,X,Z) -> pair(X,add(M,Z))
  f_2(false(),N,M,Y,X,Z) -> pair(add(M,X),Z)
  qsort(nil()) -> nil()
  qsort(add(N,X)) -> f_3(split(N,X),N,X)
  f_3(pair(Y,Z),N,X) -> append(qsort(Y),add(X,qsort(Z)))

-- SCC decomposition.

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

p1: lt#(s(X),s(Y)) -> lt#(X,Y)
p2: append#(add(N,X),Y) -> append#(X,Y)
p3: split#(N,add(M,Y)) -> f_1#(split(N,Y),N,M,Y)
p4: split#(N,add(M,Y)) -> split#(N,Y)
p5: f_1#(pair(X,Z),N,M,Y) -> f_2#(lt(N,M),N,M,Y,X,Z)
p6: f_1#(pair(X,Z),N,M,Y) -> lt#(N,M)
p7: qsort#(add(N,X)) -> f_3#(split(N,X),N,X)
p8: qsort#(add(N,X)) -> split#(N,X)
p9: f_3#(pair(Y,Z),N,X) -> append#(qsort(Y),add(X,qsort(Z)))
p10: f_3#(pair(Y,Z),N,X) -> qsort#(Y)
p11: f_3#(pair(Y,Z),N,X) -> qsort#(Z)

and R consists of:

r1: lt(|0|(),s(X)) -> true()
r2: lt(s(X),|0|()) -> false()
r3: lt(s(X),s(Y)) -> lt(X,Y)
r4: append(nil(),Y) -> Y
r5: append(add(N,X),Y) -> add(N,append(X,Y))
r6: split(N,nil()) -> pair(nil(),nil())
r7: split(N,add(M,Y)) -> f_1(split(N,Y),N,M,Y)
r8: f_1(pair(X,Z),N,M,Y) -> f_2(lt(N,M),N,M,Y,X,Z)
r9: f_2(true(),N,M,Y,X,Z) -> pair(X,add(M,Z))
r10: f_2(false(),N,M,Y,X,Z) -> pair(add(M,X),Z)
r11: qsort(nil()) -> nil()
r12: qsort(add(N,X)) -> f_3(split(N,X),N,X)
r13: f_3(pair(Y,Z),N,X) -> append(qsort(Y),add(X,qsort(Z)))

The estimated dependency graph contains the following SCCs:

  {p7, p10, p11}
  {p4}
  {p1}
  {p2}


-- Reduction pair.

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

p1: f_3#(pair(Y,Z),N,X) -> qsort#(Z)
p2: qsort#(add(N,X)) -> f_3#(split(N,X),N,X)
p3: f_3#(pair(Y,Z),N,X) -> qsort#(Y)

and R consists of:

r1: lt(|0|(),s(X)) -> true()
r2: lt(s(X),|0|()) -> false()
r3: lt(s(X),s(Y)) -> lt(X,Y)
r4: append(nil(),Y) -> Y
r5: append(add(N,X),Y) -> add(N,append(X,Y))
r6: split(N,nil()) -> pair(nil(),nil())
r7: split(N,add(M,Y)) -> f_1(split(N,Y),N,M,Y)
r8: f_1(pair(X,Z),N,M,Y) -> f_2(lt(N,M),N,M,Y,X,Z)
r9: f_2(true(),N,M,Y,X,Z) -> pair(X,add(M,Z))
r10: f_2(false(),N,M,Y,X,Z) -> pair(add(M,X),Z)
r11: qsort(nil()) -> nil()
r12: qsort(add(N,X)) -> f_3(split(N,X),N,X)
r13: f_3(pair(Y,Z),N,X) -> append(qsort(Y),add(X,qsort(Z)))

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f_3#_A(x1,x2,x3) = max{x1 + 1, x2 + 1, x3 + 20}
          pair_A(x1,x2) = max{29, x1 + 7, x2}
          qsort#_A(x1) = x1
          add_A(x1,x2) = max{x1 + 53, x2 + 21}
          split_A(x1,x2) = x2 + 19
          lt_A(x1,x2) = max{27, x1 + 1, x2 + 2}
          |0|_A = 27
          s_A(x1) = max{30, x1 + 29}
          true_A = 1
          false_A = 28
          f_2_A(x1,x2,x3,x4,x5,x6) = max{x3 + 60, x5 + 28, x6 + 21}
          f_1_A(x1,x2,x3,x4) = max{x1 + 21, x3 + 60}
          nil_A = 10
    
      precedence:
      
        false > true > split > f_1 > add = f_2 > pair > |0| > f_3# > qsort# > lt > s = nil
      
      partial status:
    
        pi(f_3#) = [1, 2]
        pi(pair) = [2]
        pi(qsort#) = [1]
        pi(add) = [1]
        pi(split) = [2]
        pi(lt) = [1]
        pi(|0|) = []
        pi(s) = [1]
        pi(true) = []
        pi(false) = []
        pi(f_2) = [3, 5]
        pi(f_1) = [1]
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f_3#_A(x1,x2,x3) = max{67, x1 + 30, x2 + 40}
          pair_A(x1,x2) = max{36, x2 + 20}
          qsort#_A(x1) = 34
          add_A(x1,x2) = 64
          split_A(x1,x2) = x2 + 37
          lt_A(x1,x2) = 79
          |0|_A = 46
          s_A(x1) = max{45, x1 + 10}
          true_A = 46
          false_A = 0
          f_2_A(x1,x2,x3,x4,x5,x6) = 36
          f_1_A(x1,x2,x3,x4) = 36
          nil_A = 37
    
      precedence:
      
        lt = s > f_3# = qsort# > split > add = true = false = f_1 > f_2 > pair = |0| = nil
      
      partial status:
    
        pi(f_3#) = [1, 2]
        pi(pair) = [2]
        pi(qsort#) = []
        pi(add) = []
        pi(split) = []
        pi(lt) = []
        pi(|0|) = []
        pi(s) = []
        pi(true) = []
        pi(false) = []
        pi(f_2) = []
        pi(f_1) = []
        pi(nil) = []
    

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: f_3#(pair(Y,Z),N,X) -> qsort#(Y)

and R consists of:

r1: lt(|0|(),s(X)) -> true()
r2: lt(s(X),|0|()) -> false()
r3: lt(s(X),s(Y)) -> lt(X,Y)
r4: append(nil(),Y) -> Y
r5: append(add(N,X),Y) -> add(N,append(X,Y))
r6: split(N,nil()) -> pair(nil(),nil())
r7: split(N,add(M,Y)) -> f_1(split(N,Y),N,M,Y)
r8: f_1(pair(X,Z),N,M,Y) -> f_2(lt(N,M),N,M,Y,X,Z)
r9: f_2(true(),N,M,Y,X,Z) -> pair(X,add(M,Z))
r10: f_2(false(),N,M,Y,X,Z) -> pair(add(M,X),Z)
r11: qsort(nil()) -> nil()
r12: qsort(add(N,X)) -> f_3(split(N,X),N,X)
r13: f_3(pair(Y,Z),N,X) -> append(qsort(Y),add(X,qsort(Z)))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: split#(N,add(M,Y)) -> split#(N,Y)

and R consists of:

r1: lt(|0|(),s(X)) -> true()
r2: lt(s(X),|0|()) -> false()
r3: lt(s(X),s(Y)) -> lt(X,Y)
r4: append(nil(),Y) -> Y
r5: append(add(N,X),Y) -> add(N,append(X,Y))
r6: split(N,nil()) -> pair(nil(),nil())
r7: split(N,add(M,Y)) -> f_1(split(N,Y),N,M,Y)
r8: f_1(pair(X,Z),N,M,Y) -> f_2(lt(N,M),N,M,Y,X,Z)
r9: f_2(true(),N,M,Y,X,Z) -> pair(X,add(M,Z))
r10: f_2(false(),N,M,Y,X,Z) -> pair(add(M,X),Z)
r11: qsort(nil()) -> nil()
r12: qsort(add(N,X)) -> f_3(split(N,X),N,X)
r13: f_3(pair(Y,Z),N,X) -> append(qsort(Y),add(X,qsort(Z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          split#_A(x1,x2) = max{0, x2 - 2}
          add_A(x1,x2) = max{3, x1, x2 + 1}
    
      precedence:
      
        split# = add
      
      partial status:
    
        pi(split#) = []
        pi(add) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          split#_A(x1,x2) = 0
          add_A(x1,x2) = x2
    
      precedence:
      
        split# = add
      
      partial status:
    
        pi(split#) = []
        pi(add) = [2]
    

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

and R consists of:

r1: lt(|0|(),s(X)) -> true()
r2: lt(s(X),|0|()) -> false()
r3: lt(s(X),s(Y)) -> lt(X,Y)
r4: append(nil(),Y) -> Y
r5: append(add(N,X),Y) -> add(N,append(X,Y))
r6: split(N,nil()) -> pair(nil(),nil())
r7: split(N,add(M,Y)) -> f_1(split(N,Y),N,M,Y)
r8: f_1(pair(X,Z),N,M,Y) -> f_2(lt(N,M),N,M,Y,X,Z)
r9: f_2(true(),N,M,Y,X,Z) -> pair(X,add(M,Z))
r10: f_2(false(),N,M,Y,X,Z) -> pair(add(M,X),Z)
r11: qsort(nil()) -> nil()
r12: qsort(add(N,X)) -> f_3(split(N,X),N,X)
r13: f_3(pair(Y,Z),N,X) -> append(qsort(Y),add(X,qsort(Z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          lt#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        lt# = s
      
      partial status:
    
        pi(lt#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          lt#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        lt# = s
      
      partial status:
    
        pi(lt#) = []
        pi(s) = [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: append#(add(N,X),Y) -> append#(X,Y)

and R consists of:

r1: lt(|0|(),s(X)) -> true()
r2: lt(s(X),|0|()) -> false()
r3: lt(s(X),s(Y)) -> lt(X,Y)
r4: append(nil(),Y) -> Y
r5: append(add(N,X),Y) -> add(N,append(X,Y))
r6: split(N,nil()) -> pair(nil(),nil())
r7: split(N,add(M,Y)) -> f_1(split(N,Y),N,M,Y)
r8: f_1(pair(X,Z),N,M,Y) -> f_2(lt(N,M),N,M,Y,X,Z)
r9: f_2(true(),N,M,Y,X,Z) -> pair(X,add(M,Z))
r10: f_2(false(),N,M,Y,X,Z) -> pair(add(M,X),Z)
r11: qsort(nil()) -> nil()
r12: qsort(add(N,X)) -> f_3(split(N,X),N,X)
r13: f_3(pair(Y,Z),N,X) -> append(qsort(Y),add(X,qsort(Z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          append#_A(x1,x2) = max{x1 + 1, x2}
          add_A(x1,x2) = max{x1 - 1, x2}
    
      precedence:
      
        append# = add
      
      partial status:
    
        pi(append#) = [1, 2]
        pi(add) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          append#_A(x1,x2) = max{x1 - 1, x2 + 1}
          add_A(x1,x2) = x2
    
      precedence:
      
        append# = add
      
      partial status:
    
        pi(append#) = []
        pi(add) = [2]
    

The next rules are strictly ordered:

  p1

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