(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))
s(log(0)) → s(0)
log(s(x)) → s(log(half(s(x))))
Q is empty.
(1) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0'(half(x)) → 0'(x)
0'(s(half(x))) → 0'(x)
s(s(half(x))) → half(s(x))
0'(log(s(x))) → 0'(s(x))
s(log(x)) → s(half(log(s(x))))
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0'(x1)) = x1
POL(half(x1)) = x1
POL(log(x1)) = 1 + x1
POL(s(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
0'(log(s(x))) → 0'(s(x))
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0'(half(x)) → 0'(x)
0'(s(half(x))) → 0'(x)
s(s(half(x))) → half(s(x))
s(log(x)) → s(half(log(s(x))))
Q is empty.
(5) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
s(s(half(x))) → half(s(x))
s(log(x)) → s(half(log(s(x))))
The TRS R 2 is
0'(half(x)) → 0'(x)
0'(s(half(x))) → 0'(x)
The signature Sigma is {
0'}
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0'(half(x)) → 0'(x)
0'(s(half(x))) → 0'(x)
s(s(half(x))) → half(s(x))
s(log(x)) → s(half(log(s(x))))
The set Q consists of the following terms:
0'(half(x0))
0'(s(half(x0)))
s(s(half(x0)))
s(log(x0))
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
0'1(half(x)) → 0'1(x)
0'1(s(half(x))) → 0'1(x)
S(s(half(x))) → S(x)
S(log(x)) → S(half(log(s(x))))
S(log(x)) → S(x)
The TRS R consists of the following rules:
0'(half(x)) → 0'(x)
0'(s(half(x))) → 0'(x)
s(s(half(x))) → half(s(x))
s(log(x)) → s(half(log(s(x))))
The set Q consists of the following terms:
0'(half(x0))
0'(s(half(x0)))
s(s(half(x0)))
s(log(x0))
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(10) Complex Obligation (AND)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(log(x)) → S(x)
S(s(half(x))) → S(x)
The TRS R consists of the following rules:
0'(half(x)) → 0'(x)
0'(s(half(x))) → 0'(x)
s(s(half(x))) → half(s(x))
s(log(x)) → s(half(log(s(x))))
The set Q consists of the following terms:
0'(half(x0))
0'(s(half(x0)))
s(s(half(x0)))
s(log(x0))
We have to consider all minimal (P,Q,R)-chains.
(12) 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.
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(log(x)) → S(x)
S(s(half(x))) → S(x)
R is empty.
The set Q consists of the following terms:
0'(half(x0))
0'(s(half(x0)))
s(s(half(x0)))
s(log(x0))
We have to consider all minimal (P,Q,R)-chains.
(14) 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].
0'(half(x0))
0'(s(half(x0)))
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(log(x)) → S(x)
S(s(half(x))) → S(x)
R is empty.
The set Q consists of the following terms:
s(s(half(x0)))
s(log(x0))
We have to consider all minimal (P,Q,R)-chains.
(16) 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:
- S(log(x)) → S(x)
The graph contains the following edges 1 > 1
- S(s(half(x))) → S(x)
The graph contains the following edges 1 > 1
(17) YES
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
0'1(s(half(x))) → 0'1(x)
0'1(half(x)) → 0'1(x)
The TRS R consists of the following rules:
0'(half(x)) → 0'(x)
0'(s(half(x))) → 0'(x)
s(s(half(x))) → half(s(x))
s(log(x)) → s(half(log(s(x))))
The set Q consists of the following terms:
0'(half(x0))
0'(s(half(x0)))
s(s(half(x0)))
s(log(x0))
We have to consider all minimal (P,Q,R)-chains.
(19) 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.
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
0'1(s(half(x))) → 0'1(x)
0'1(half(x)) → 0'1(x)
R is empty.
The set Q consists of the following terms:
0'(half(x0))
0'(s(half(x0)))
s(s(half(x0)))
s(log(x0))
We have to consider all minimal (P,Q,R)-chains.
(21) 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].
0'(half(x0))
0'(s(half(x0)))
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
0'1(s(half(x))) → 0'1(x)
0'1(half(x)) → 0'1(x)
R is empty.
The set Q consists of the following terms:
s(s(half(x0)))
s(log(x0))
We have to consider all minimal (P,Q,R)-chains.
(23) 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:
- 0'1(s(half(x))) → 0'1(x)
The graph contains the following edges 1 > 1
- 0'1(half(x)) → 0'1(x)
The graph contains the following edges 1 > 1
(24) YES