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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f_3#_A(x1,x2,x3) = x1 + (1,6)
          pair_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (2,2)
          qsort#_A(x1) = x1 + (1,1)
          add_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (6,10)
          split_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (4,3)
          lt_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (11,11)
          |0|_A() = (10,4)
          s_A(x1) = ((1,0),(0,0)) x1 + (10,0)
          true_A() = (1,1)
          false_A() = (9,3)
          f_2_A(x1,x2,x3,x4,x5,x6) = ((1,0),(1,0)) x3 + ((1,0),(0,0)) x5 + ((1,0),(1,0)) x6 + (8,9)
          f_1_A(x1,x2,x3,x4) = x1 + ((1,0),(1,1)) x3 + (6,8)
          nil_A() = (1,4)
  
    precedence:
    
      lt = |0| = true > split = f_1 > add = s = false > f_3# = qsort# = f_2 > pair > nil
    
    partial status:
  
      pi(f_3#) = [1]
      pi(pair) = []
      pi(qsort#) = [1]
      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:

  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#(Z)
p2: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          split#_A(x1,x2) = ((1,0),(1,0)) x2 + (1,0)
          add_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (1,0)
  
    precedence:
    
      add > split#
    
    partial status:
  
      pi(split#) = []
      pi(add) = []

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          lt#_A(x1,x2) = x1
          s_A(x1) = x1 + (1,1)
  
    precedence:
    
      s > lt#
    
    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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          append#_A(x1,x2) = ((1,0),(0,0)) x1 + (1,1)
          add_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (2,2)
  
    precedence:
    
      append# = add
    
    partial status:
  
      pi(append#) = []
      pi(add) = [1]

The next rules are strictly ordered:

  p1

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