(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) TRUE
(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) TRUE
(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: Polynomial interpretation [POLO]:
POL(.(x1, x2)) = 1 + x1 + x2
POL(BUBBLE(x1)) = x1
(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) TRUE
(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].
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: Matrix interpretation [MATRO]:
| POL(.(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
| POL(if(x1, x2, x3)) = | | + | | · | x1 | + | | · | x2 | + | | · | x3 |
| POL(<=(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))
butlast(nil) → nil
butlast(.(x, nil)) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(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) TRUE