(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
1(0(x)) → 0(0(0(1(x))))
0(1(x)) → 1(x)
1(1(x)) → 0(0(0(0(x))))
0(0(x)) → 0(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(1(x)) → 1(0(0(0(x))))
1(0(x)) → 1(x)
1(1(x)) → 0(0(0(0(x))))
0(0(x)) → 0(x)
Q is empty.
(3) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(1(x)) → 1(0(0(0(x))))
1(0(x)) → 1(x)
1(1(x)) → 0(0(0(0(x))))
0(0(x)) → 0(x)
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = x1
POL(1(x1)) = 1 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
1(1(x)) → 0(0(0(0(x))))
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
1(0(x)) → 0(0(0(1(x))))
0(1(x)) → 1(x)
0(0(x)) → 0(x)
Q is empty.
(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:
11(0(x)) → 01(0(0(1(x))))
11(0(x)) → 01(0(1(x)))
11(0(x)) → 01(1(x))
11(0(x)) → 11(x)
The TRS R consists of the following rules:
1(0(x)) → 0(0(0(1(x))))
0(1(x)) → 1(x)
0(0(x)) → 0(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
11(0(x)) → 11(x)
The TRS R consists of the following rules:
1(0(x)) → 0(0(0(1(x))))
0(1(x)) → 1(x)
0(0(x)) → 0(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
11(0(x)) → 11(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
11(0(x)) → 11(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) UsableRulesReductionPairsProof (EQUIVALENT transformation)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
11(0(x)) → 11(x)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(0(x1)) = 2·x1
POL(11(x1)) = 2·x1
(16) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(18) TRUE