YES Termination w.r.t. Q proof of SK90_4.61.ari

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 Overlay + Local Confluence (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 MRRProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QReductionProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QDPOrderProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 PisEmptyProof (⇔, 0 ms)
        ↳38 YES

(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) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

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

R is empty.
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) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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)))

(11) Obligation:

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

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

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • BUTLAST(.(x, .(y, z))) → BUTLAST(.(y, z))
    The graph contains the following edges 1 > 1

(13) YES

(14) 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.

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

LAST(.(x, .(y, z))) → LAST(.(y, z))

R is empty.
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.

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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)))

(18) Obligation:

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

LAST(.(x, .(y, z))) → LAST(.(y, z))

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • LAST(.(x, .(y, z))) → LAST(.(y, z))
    The graph contains the following edges 1 > 1

(20) YES

(21) 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.

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) 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))

R is empty.
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.

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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)))

(25) 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))

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

(26) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: Knuth-Bendix order [KBO] with precedence:
.2 > BUBBLE1

and weight map:

BUBBLE_1=1
._2=0

The variable weight is 1

(27) Obligation:

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) YES

(30) 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.

(31) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(32) 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:

bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(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.

(33) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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

(34) 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:

bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(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:

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

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


BSORT(.(x, y)) → BSORT(butlast(bubble(.(x, y))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( BSORT(x1) ) = 2x1

POL( butlast(x1) ) = max{0, x1 - 2}

POL( bubble(x1) ) = x1 + 1

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

POL( nil ) = 0

POL( if(x1, ..., x3) ) = x3

POL( <=(x1, x2) ) = 0


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

(36) Obligation:

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

bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(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:

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

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

(37) PisEmptyProof (EQUIVALENT transformation)

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

(38) YES