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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f_3#_A(x1,x2,x3) = x1 + ((0,1),(0,0)) x2
        pair_A(x1,x2) = x1 + x2 + (1,2)
        qsort#_A(x1) = x1
        add_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,1),(1,1)) x2 + (4,2)
        split_A(x1,x2) = ((1,1),(1,1)) x2 + (3,1)
        lt_A(x1,x2) = ((0,1),(1,0)) x2 + (1,1)
        |0|_A() = (1,0)
        s_A(x1) = ((1,1),(0,1)) x1 + (1,1)
        true_A() = (0,1)
        false_A() = (0,0)
        f_2_A(x1,x2,x3,x4,x5,x6) = ((0,0),(1,0)) x1 + ((0,1),(0,0)) x3 + ((1,1),(1,1)) x5 + ((1,1),(1,1)) x6 + (6,4)
        f_1_A(x1,x2,x3,x4) = ((1,1),(1,1)) x1 + ((0,1),(0,1)) x3 + (4,2)
        nil_A() = (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f_3#_A(x1,x2,x3) = (4,5)
        pair_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,0)) x2 + (1,1)
        qsort#_A(x1) = ((1,0),(1,1)) x1
        add_A(x1,x2) = ((1,1),(1,1)) x2 + (3,1)
        split_A(x1,x2) = x2 + (2,1)
        lt_A(x1,x2) = (1,1)
        |0|_A() = (1,1)
        s_A(x1) = (1,1)
        true_A() = (2,2)
        false_A() = (2,2)
        f_2_A(x1,x2,x3,x4,x5,x6) = (6,3)
        f_1_A(x1,x2,x3,x4) = ((1,1),(1,1)) x1 + (3,0)
        nil_A() = (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3

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: 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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        split#_A(x1,x2) = x1 + ((1,1),(1,1)) x2
        add_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        split#_A(x1,x2) = x1 + ((0,1),(1,0)) x2
        add_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,0),(1,0)) x2 + (1,1)
    

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: 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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        lt#_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,0)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        lt#_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(1,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    

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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        append#_A(x1,x2) = ((1,1),(1,1)) x1 + x2
        add_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        append#_A(x1,x2) = ((0,1),(1,0)) x1 + x2
        add_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,0),(1,0)) x2 + (1,1)
    

The next rules are strictly ordered:

  p1

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