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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        bsort#_A(x1) = max{12, x1 + 3}
        ._A(x1,x2) = x2 + 22
        butlast_A(x1) = max{13, x1 - 9}
        bubble_A(x1) = max{27, x1 + 4}
        nil_A = 1
        if_A(x1,x2,x3) = max{48, x1 - 55, x2 - 1, x3 - 1}
        <=_A(x1,x2) = 49
  
    precedence:
    
      bsort# = . = butlast = bubble = nil = if = <=
    
    partial status:
  
      pi(bsort#) = [1]
      pi(.) = []
      pi(butlast) = []
      pi(bubble) = []
      pi(nil) = []
      pi(if) = []
      pi(<=) = []

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 reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        last#_A(x1) = x1 + 6
        ._A(x1,x2) = max{x1 - 2, x2 + 1}
  
    precedence:
    
      last# = .
    
    partial status:
  
      pi(last#) = []
      pi(.) = [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: 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 reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        bubble#_A(x1) = x1 + 6
        ._A(x1,x2) = max{x1 - 7, x2 + 2}
  
    precedence:
    
      bubble# = .
    
    partial status:
  
      pi(bubble#) = []
      pi(.) = [2]

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

  {p1}


-- Reduction pair.

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

p1: bubble#(.(x,.(y,z))) -> bubble#(.(x,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 reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        bubble#_A(x1) = x1
        ._A(x1,x2) = max{x1, x2}
  
    precedence:
    
      . > bubble#
    
    partial status:
  
      pi(bubble#) = [1]
      pi(.) = [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: 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 reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        butlast#_A(x1) = x1 + 6
        ._A(x1,x2) = max{x1 - 2, x2 + 1}
  
    precedence:
    
      butlast# = .
    
    partial status:
  
      pi(butlast#) = []
      pi(.) = [2]

The next rules are strictly ordered:

  p1

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