(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))
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:
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
conv(0)
conv(s(x0))
(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:
HALF(s(s(x))) → HALF(x)
LASTBIT(s(s(x))) → LASTBIT(x)
CONV(s(x)) → CONV(half(s(x)))
CONV(s(x)) → HALF(s(x))
CONV(s(x)) → LASTBIT(s(x))
The TRS R consists of the following rules:
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
conv(0)
conv(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LASTBIT(s(s(x))) → LASTBIT(x)
The TRS R consists of the following rules:
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
conv(0)
conv(s(x0))
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:
LASTBIT(s(s(x))) → LASTBIT(x)
R is empty.
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
conv(0)
conv(s(x0))
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].
half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
conv(0)
conv(s(x0))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LASTBIT(s(s(x))) → LASTBIT(x)
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:
- LASTBIT(s(s(x))) → LASTBIT(x)
The graph contains the following edges 1 > 1
(13) TRUE
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x))) → HALF(x)
The TRS R consists of the following rules:
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
conv(0)
conv(s(x0))
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:
HALF(s(s(x))) → HALF(x)
R is empty.
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
conv(0)
conv(s(x0))
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].
half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
conv(0)
conv(s(x0))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x))) → HALF(x)
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:
- HALF(s(s(x))) → HALF(x)
The graph contains the following edges 1 > 1
(20) TRUE
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CONV(s(x)) → CONV(half(s(x)))
The TRS R consists of the following rules:
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
conv(0) → cons(nil, 0)
conv(s(x)) → cons(conv(half(s(x))), lastbit(s(x)))
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
conv(0)
conv(s(x0))
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:
CONV(s(x)) → CONV(half(s(x)))
The TRS R consists of the following rules:
half(s(0)) → 0
half(s(s(x))) → s(half(x))
half(0) → 0
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
conv(0)
conv(s(x0))
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].
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
conv(0)
conv(s(x0))
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CONV(s(x)) → CONV(half(s(x)))
The TRS R consists of the following rules:
half(s(0)) → 0
half(s(s(x))) → s(half(x))
half(0) → 0
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
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 rules of the TRS R:
half(s(0)) → 0
half(s(s(x))) → s(half(x))
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(CONV(x1)) = x1
POL(half(x1)) = x1
POL(s(x1)) = 1 + x1
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CONV(s(x)) → CONV(half(s(x)))
The TRS R consists of the following rules:
half(0) → 0
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
(28) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(29) TRUE