YES

We show the termination of the TRS R:

  bsort(nil()) -> nil()
  bsort(.(x,y)) -> last(.(bubble(.(x,y)),bsort(butlast(bubble(.(x,y))))))
  bubble(nil()) -> nil()
  bubble(.(x,nil())) -> .(x,nil())
  bubble(.(x,.(y,z))) -> if(<=(x,y),.(y,bubble(.(x,z))),.(x,bubble(.(y,z))))
  last(nil()) -> |0|()
  last(.(x,nil())) -> x
  last(.(x,.(y,z))) -> last(.(y,z))
  butlast(nil()) -> nil()
  butlast(.(x,nil())) -> nil()
  butlast(.(x,.(y,z))) -> .(x,butlast(.(y,z)))

-- SCC decomposition.

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

p1: bsort#(.(x,y)) -> last#(.(bubble(.(x,y)),bsort(butlast(bubble(.(x,y))))))
p2: bsort#(.(x,y)) -> bubble#(.(x,y))
p3: bsort#(.(x,y)) -> bsort#(butlast(bubble(.(x,y))))
p4: bsort#(.(x,y)) -> butlast#(bubble(.(x,y)))
p5: bubble#(.(x,.(y,z))) -> bubble#(.(x,z))
p6: bubble#(.(x,.(y,z))) -> bubble#(.(y,z))
p7: last#(.(x,.(y,z))) -> last#(.(y,z))
p8: butlast#(.(x,.(y,z))) -> butlast#(.(y,z))

and R consists of:

r1: bsort(nil()) -> nil()
r2: bsort(.(x,y)) -> last(.(bubble(.(x,y)),bsort(butlast(bubble(.(x,y))))))
r3: bubble(nil()) -> nil()
r4: bubble(.(x,nil())) -> .(x,nil())
r5: bubble(.(x,.(y,z))) -> if(<=(x,y),.(y,bubble(.(x,z))),.(x,bubble(.(y,z))))
r6: last(nil()) -> |0|()
r7: last(.(x,nil())) -> x
r8: last(.(x,.(y,z))) -> last(.(y,z))
r9: butlast(nil()) -> nil()
r10: butlast(.(x,nil())) -> nil()
r11: butlast(.(x,.(y,z))) -> .(x,butlast(.(y,z)))

The estimated dependency graph contains the following SCCs:

  {p3}
  {p7}
  {p5, p6}
  {p8}


-- Reduction pair.

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

p1: bsort#(.(x,y)) -> bsort#(butlast(bubble(.(x,y))))

and R consists of:

r1: bsort(nil()) -> nil()
r2: bsort(.(x,y)) -> last(.(bubble(.(x,y)),bsort(butlast(bubble(.(x,y))))))
r3: bubble(nil()) -> nil()
r4: bubble(.(x,nil())) -> .(x,nil())
r5: bubble(.(x,.(y,z))) -> if(<=(x,y),.(y,bubble(.(x,z))),.(x,bubble(.(y,z))))
r6: last(nil()) -> |0|()
r7: last(.(x,nil())) -> x
r8: last(.(x,.(y,z))) -> last(.(y,z))
r9: butlast(nil()) -> nil()
r10: butlast(.(x,nil())) -> nil()
r11: butlast(.(x,.(y,z))) -> .(x,butlast(.(y,z)))

The set of usable rules consists of

  r4, r5, r9, r10, r11

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      bsort#_A(x1) = ((0,1),(1,0)) x1
      ._A(x1,x2) = x2 + (1,2)
      butlast_A(x1) = ((1,0),(1,0)) x1
      bubble_A(x1) = x1 + (1,4)
      nil_A() = (0,0)
      if_A(x1,x2,x3) = ((0,0),(1,0)) x2 + ((0,0),(1,1)) x3 + (1,0)
      <=_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2

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: last#(.(x,.(y,z))) -> last#(.(y,z))

and R consists of:

r1: bsort(nil()) -> nil()
r2: bsort(.(x,y)) -> last(.(bubble(.(x,y)),bsort(butlast(bubble(.(x,y))))))
r3: bubble(nil()) -> nil()
r4: bubble(.(x,nil())) -> .(x,nil())
r5: bubble(.(x,.(y,z))) -> if(<=(x,y),.(y,bubble(.(x,z))),.(x,bubble(.(y,z))))
r6: last(nil()) -> |0|()
r7: last(.(x,nil())) -> x
r8: last(.(x,.(y,z))) -> last(.(y,z))
r9: butlast(nil()) -> nil()
r10: butlast(.(x,nil())) -> nil()
r11: butlast(.(x,.(y,z))) -> .(x,butlast(.(y,z)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

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: bubble#(.(x,.(y,z))) -> bubble#(.(x,z))
p2: bubble#(.(x,.(y,z))) -> bubble#(.(y,z))

and R consists of:

r1: bsort(nil()) -> nil()
r2: bsort(.(x,y)) -> last(.(bubble(.(x,y)),bsort(butlast(bubble(.(x,y))))))
r3: bubble(nil()) -> nil()
r4: bubble(.(x,nil())) -> .(x,nil())
r5: bubble(.(x,.(y,z))) -> if(<=(x,y),.(y,bubble(.(x,z))),.(x,bubble(.(y,z))))
r6: last(nil()) -> |0|()
r7: last(.(x,nil())) -> x
r8: last(.(x,.(y,z))) -> last(.(y,z))
r9: butlast(nil()) -> nil()
r10: butlast(.(x,nil())) -> nil()
r11: butlast(.(x,.(y,z))) -> .(x,butlast(.(y,z)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

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: butlast#(.(x,.(y,z))) -> butlast#(.(y,z))

and R consists of:

r1: bsort(nil()) -> nil()
r2: bsort(.(x,y)) -> last(.(bubble(.(x,y)),bsort(butlast(bubble(.(x,y))))))
r3: bubble(nil()) -> nil()
r4: bubble(.(x,nil())) -> .(x,nil())
r5: bubble(.(x,.(y,z))) -> if(<=(x,y),.(y,bubble(.(x,z))),.(x,bubble(.(y,z))))
r6: last(nil()) -> |0|()
r7: last(.(x,nil())) -> x
r8: last(.(x,.(y,z))) -> last(.(y,z))
r9: butlast(nil()) -> nil()
r10: butlast(.(x,nil())) -> nil()
r11: butlast(.(x,.(y,z))) -> .(x,butlast(.(y,z)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

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