Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

BUBBLE1(.2(x, .2(y, z))) -> BUBBLE1(.2(x, z))
BSORT1(.2(x, y)) -> BSORT1(butlast1(bubble1(.2(x, y))))
BSORT1(.2(x, y)) -> LAST1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
LAST1(.2(x, .2(y, z))) -> LAST1(.2(y, z))
BUTLAST1(.2(x, .2(y, z))) -> BUTLAST1(.2(y, z))
BSORT1(.2(x, y)) -> BUTLAST1(bubble1(.2(x, y)))
BUBBLE1(.2(x, .2(y, z))) -> BUBBLE1(.2(y, z))
BSORT1(.2(x, y)) -> BUBBLE1(.2(x, y))

The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

BUBBLE1(.2(x, .2(y, z))) -> BUBBLE1(.2(x, z))
BSORT1(.2(x, y)) -> BSORT1(butlast1(bubble1(.2(x, y))))
BSORT1(.2(x, y)) -> LAST1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
LAST1(.2(x, .2(y, z))) -> LAST1(.2(y, z))
BUTLAST1(.2(x, .2(y, z))) -> BUTLAST1(.2(y, z))
BSORT1(.2(x, y)) -> BUTLAST1(bubble1(.2(x, y)))
BUBBLE1(.2(x, .2(y, z))) -> BUBBLE1(.2(y, z))
BSORT1(.2(x, y)) -> BUBBLE1(.2(x, y))

The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

BUTLAST1(.2(x, .2(y, z))) -> BUTLAST1(.2(y, z))

The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


BUTLAST1(.2(x, .2(y, z))) -> BUTLAST1(.2(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( BUTLAST1(x1) ) = max{0, x1 - 1}


POL( .2(x1, x2) ) = x2 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

LAST1(.2(x, .2(y, z))) -> LAST1(.2(y, z))

The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


LAST1(.2(x, .2(y, z))) -> LAST1(.2(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( LAST1(x1) ) = max{0, x1 - 1}


POL( .2(x1, x2) ) = x2 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

BUBBLE1(.2(x, .2(y, z))) -> BUBBLE1(.2(x, z))
BUBBLE1(.2(x, .2(y, z))) -> BUBBLE1(.2(y, z))

The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


BUBBLE1(.2(x, .2(y, z))) -> BUBBLE1(.2(x, z))
BUBBLE1(.2(x, .2(y, z))) -> BUBBLE1(.2(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( BUBBLE1(x1) ) = max{0, x1 - 1}


POL( .2(x1, x2) ) = x2 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

BSORT1(.2(x, y)) -> BSORT1(butlast1(bubble1(.2(x, y))))

The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


BSORT1(.2(x, y)) -> BSORT1(butlast1(bubble1(.2(x, y))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( BSORT1(x1) ) = x1


POL( .2(x1, x2) ) = x2 + 1


POL( butlast1(x1) ) = max{0, x1 - 1}


POL( bubble1(x1) ) = 1


POL( nil ) = 0


POL( if3(x1, ..., x3) ) = 0



The following usable rules [14] were oriented:

bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
butlast1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))
butlast1(.2(x, nil)) -> nil



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

bsort1(nil) -> nil
bsort1(.2(x, y)) -> last1(.2(bubble1(.2(x, y)), bsort1(butlast1(bubble1(.2(x, y))))))
bubble1(nil) -> nil
bubble1(.2(x, nil)) -> .2(x, nil)
bubble1(.2(x, .2(y, z))) -> if3(<=2(x, y), .2(y, bubble1(.2(x, z))), .2(x, bubble1(.2(y, z))))
last1(nil) -> 0
last1(.2(x, nil)) -> x
last1(.2(x, .2(y, z))) -> last1(.2(y, z))
butlast1(nil) -> nil
butlast1(.2(x, nil)) -> nil
butlast1(.2(x, .2(y, z))) -> .2(x, butlast1(.2(y, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.