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:
  
      max/plus interpretations on natural numbers:
      
        f_3#_A(x1,x2,x3) = max{x1 + 7, x2 + 21, x3 + 6}
        pair_A(x1,x2) = max{x1 + 1, x2 + 6}
        qsort#_A(x1) = x1 + 7
        add_A(x1,x2) = max{x1 + 21, x2 + 22}
        split_A(x1,x2) = x2 + 15
        lt_A(x1,x2) = max{30, x1 - 31, x2 + 3}
        |0|_A = 30
        s_A(x1) = x1 + 30
        true_A = 33
        false_A = 29
        f_2_A(x1,x2,x3,x4,x5,x6) = max{x3 + 28, x5 + 23, x6 + 28}
        f_1_A(x1,x2,x3,x4) = max{x1 + 22, x3 + 28}
        nil_A = 3
  
    precedence:
    
      add > f_3# > pair = qsort# = split = lt = |0| = s = true = false = f_2 = f_1 = nil
    
    partial status:
  
      pi(f_3#) = [1, 2, 3]
      pi(pair) = []
      pi(qsort#) = [1]
      pi(add) = [2]
      pi(split) = []
      pi(lt) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(true) = []
      pi(false) = []
      pi(f_2) = []
      pi(f_1) = []
      pi(nil) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: qsort#(add(N,X)) -> f_3#(split(N,X),N,X)
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:

  {p1, p2}


-- Reduction pair.

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

p1: qsort#(add(N,X)) -> f_3#(split(N,X),N,X)
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 set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        qsort#_A(x1) = max{5, x1}
        add_A(x1,x2) = max{x1 + 5, x2 + 10}
        f_3#_A(x1,x2,x3) = max{x1 - 1, x2 + 5, x3 + 10}
        split_A(x1,x2) = x2 + 11
        pair_A(x1,x2) = max{x1 + 6, x2 + 6}
        lt_A(x1,x2) = max{3, x2 - 12}
        |0|_A = 22
        s_A(x1) = x1 + 11
        true_A = 0
        false_A = 10
        f_2_A(x1,x2,x3,x4,x5,x6) = max{x1 - 5, x3 + 13, x5 + 16, x6 + 16}
        f_1_A(x1,x2,x3,x4) = max{x1 + 10, x3 + 13}
        nil_A = 2
  
    precedence:
    
      qsort# = add = f_3# = split = pair = lt = |0| = s = true = false = f_2 = f_1 = nil
    
    partial status:
  
      pi(qsort#) = [1]
      pi(add) = [1]
      pi(f_3#) = []
      pi(split) = []
      pi(pair) = []
      pi(lt) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(true) = []
      pi(false) = []
      pi(f_2) = []
      pi(f_1) = []
      pi(nil) = []

The next rules are strictly ordered:

  p1

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        split#_A(x1,x2) = max{0, x2 - 2}
        add_A(x1,x2) = max{x1 + 3, x2 + 1}
  
    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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        lt#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      lt# = s
    
    partial status:
  
      pi(lt#) = []
      pi(s) = []

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:
  
      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#) = [1]
      pi(add) = [2]

The next rules are strictly ordered:

  p1

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