Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

bsort1(nil)
bsort1(.2(x0, x1))
bubble1(nil)
bubble1(.2(x0, nil))
bubble1(.2(x0, .2(x1, x2)))
last1(nil)
last1(.2(x0, nil))
last1(.2(x0, .2(x1, x2)))
butlast1(nil)
butlast1(.2(x0, nil))
butlast1(.2(x0, .2(x1, x2)))


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)))

The set Q consists of the following terms:

bsort1(nil)
bsort1(.2(x0, x1))
bubble1(nil)
bubble1(.2(x0, nil))
bubble1(.2(x0, .2(x1, x2)))
last1(nil)
last1(.2(x0, nil))
last1(.2(x0, .2(x1, x2)))
butlast1(nil)
butlast1(.2(x0, nil))
butlast1(.2(x0, .2(x1, x2)))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

bsort1(nil)
bsort1(.2(x0, x1))
bubble1(nil)
bubble1(.2(x0, nil))
bubble1(.2(x0, .2(x1, x2)))
last1(nil)
last1(.2(x0, nil))
last1(.2(x0, .2(x1, x2)))
butlast1(nil)
butlast1(.2(x0, nil))
butlast1(.2(x0, .2(x1, x2)))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ 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)))

The set Q consists of the following terms:

bsort1(nil)
bsort1(.2(x0, x1))
bubble1(nil)
bubble1(.2(x0, nil))
bubble1(.2(x0, .2(x1, x2)))
last1(nil)
last1(.2(x0, nil))
last1(.2(x0, .2(x1, x2)))
butlast1(nil)
butlast1(.2(x0, nil))
butlast1(.2(x0, .2(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

BUTLAST1(.2(x, .2(y, z))) -> BUTLAST1(.2(y, z))
Used argument filtering: BUTLAST1(x1)  =  x1
.2(x1, x2)  =  .1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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)))

The set Q consists of the following terms:

bsort1(nil)
bsort1(.2(x0, x1))
bubble1(nil)
bubble1(.2(x0, nil))
bubble1(.2(x0, .2(x1, x2)))
last1(nil)
last1(.2(x0, nil))
last1(.2(x0, .2(x1, x2)))
butlast1(nil)
butlast1(.2(x0, nil))
butlast1(.2(x0, .2(x1, x2)))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ 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)))

The set Q consists of the following terms:

bsort1(nil)
bsort1(.2(x0, x1))
bubble1(nil)
bubble1(.2(x0, nil))
bubble1(.2(x0, .2(x1, x2)))
last1(nil)
last1(.2(x0, nil))
last1(.2(x0, .2(x1, x2)))
butlast1(nil)
butlast1(.2(x0, nil))
butlast1(.2(x0, .2(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LAST1(.2(x, .2(y, z))) -> LAST1(.2(y, z))
Used argument filtering: LAST1(x1)  =  x1
.2(x1, x2)  =  .1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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)))

The set Q consists of the following terms:

bsort1(nil)
bsort1(.2(x0, x1))
bubble1(nil)
bubble1(.2(x0, nil))
bubble1(.2(x0, .2(x1, x2)))
last1(nil)
last1(.2(x0, nil))
last1(.2(x0, .2(x1, x2)))
butlast1(nil)
butlast1(.2(x0, nil))
butlast1(.2(x0, .2(x1, x2)))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ 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)))

The set Q consists of the following terms:

bsort1(nil)
bsort1(.2(x0, x1))
bubble1(nil)
bubble1(.2(x0, nil))
bubble1(.2(x0, .2(x1, x2)))
last1(nil)
last1(.2(x0, nil))
last1(.2(x0, .2(x1, x2)))
butlast1(nil)
butlast1(.2(x0, nil))
butlast1(.2(x0, .2(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

BUBBLE1(.2(x, .2(y, z))) -> BUBBLE1(.2(x, z))
BUBBLE1(.2(x, .2(y, z))) -> BUBBLE1(.2(y, z))
Used argument filtering: BUBBLE1(x1)  =  x1
.2(x1, x2)  =  .1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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)))

The set Q consists of the following terms:

bsort1(nil)
bsort1(.2(x0, x1))
bubble1(nil)
bubble1(.2(x0, nil))
bubble1(.2(x0, .2(x1, x2)))
last1(nil)
last1(.2(x0, nil))
last1(.2(x0, .2(x1, x2)))
butlast1(nil)
butlast1(.2(x0, nil))
butlast1(.2(x0, .2(x1, x2)))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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)))

The set Q consists of the following terms:

bsort1(nil)
bsort1(.2(x0, x1))
bubble1(nil)
bubble1(.2(x0, nil))
bubble1(.2(x0, .2(x1, x2)))
last1(nil)
last1(.2(x0, nil))
last1(.2(x0, .2(x1, x2)))
butlast1(nil)
butlast1(.2(x0, nil))
butlast1(.2(x0, .2(x1, x2)))

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