Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

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)))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

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)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

BSORT(.(x, y)) → LAST(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
BSORT(.(x, y)) → BUBBLE(.(x, y))
BSORT(.(x, y)) → BSORT(butlast(bubble(.(x, y))))
BSORT(.(x, y)) → BUTLAST(bubble(.(x, y)))
BUBBLE(.(x, .(y, z))) → BUBBLE(.(x, z))
BUBBLE(.(x, .(y, z))) → BUBBLE(.(y, z))
LAST(.(x, .(y, z))) → LAST(.(y, z))
BUTLAST(.(x, .(y, z))) → BUTLAST(.(y, z))

The TRS R consists of the following rules:

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)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

BUTLAST(.(x, .(y, z))) → BUTLAST(.(y, z))

The TRS R consists of the following rules:

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)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

BUTLAST(.(x, .(y, z))) → BUTLAST(.(y, z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
BUTLAST(x1)  =  x1
.(x1, x2)  =  .(x2)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

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)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

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

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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

LAST(.(x, .(y, z))) → LAST(.(y, z))

The TRS R consists of the following rules:

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)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LAST(.(x, .(y, z))) → LAST(.(y, z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LAST(x1)  =  x1
.(x1, x2)  =  .(x2)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

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)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

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

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) Obligation:

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

BUBBLE(.(x, .(y, z))) → BUBBLE(.(y, z))
BUBBLE(.(x, .(y, z))) → BUBBLE(.(x, z))

The TRS R consists of the following rules:

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)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

BUBBLE(.(x, .(y, z))) → BUBBLE(.(y, z))
BUBBLE(.(x, .(y, z))) → BUBBLE(.(x, z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
BUBBLE(x1)  =  x1
.(x1, x2)  =  .(x2)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

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)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

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

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE

(22) Obligation:

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

BSORT(.(x, y)) → BSORT(butlast(bubble(.(x, y))))

The TRS R consists of the following rules:

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)))

The set Q consists of the following terms:

bsort(nil)
bsort(.(x0, x1))
bubble(nil)
bubble(.(x0, nil))
bubble(.(x0, .(x1, x2)))
last(nil)
last(.(x0, nil))
last(.(x0, .(x1, x2)))
butlast(nil)
butlast(.(x0, nil))
butlast(.(x0, .(x1, x2)))

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