(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(x)))) → b(b(b(a(x))))
b(a(x)) → a(a(a(a(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:
b(b(a(a(x)))) → a(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
Q is empty.
(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:
B(b(a(a(x)))) → A(b(b(b(x))))
B(b(a(a(x)))) → B(b(b(x)))
B(b(a(a(x)))) → B(b(x))
B(b(a(a(x)))) → B(x)
A(b(x)) → A(a(a(a(x))))
A(b(x)) → A(a(a(x)))
A(b(x)) → A(a(x))
A(b(x)) → A(x)
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) 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:
- A(b(x)) → A(x)
The graph contains the following edges 1 > 1
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(x)))) → B(b(x))
B(b(a(a(x)))) → B(b(b(x)))
B(b(a(a(x)))) → B(x)
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(b(a(a(x)))) → B(b(b(x)))
B(b(a(a(x)))) → B(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic integers [ARCTIC,STERNAGEL_THIEMANN_RTA14]:
| POL(b(x1)) = | | + | | / | -I | -I | 0A | \ |
| | | -I | -I | -1A | | |
| \ | 0A | 1A | -1A | / |
| · | x1 |
| POL(a(x1)) = | | + | | / | -1A | 0A | 0A | \ |
| | | 0A | -1A | -1A | | |
| \ | -1A | -1A | -1A | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
b(b(a(a(x)))) → a(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(x)))) → B(b(x))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) SemLabProof (SOUND transformation)
We found the following model for the rules of the TRSs R and P.
Interpretation over the domain with elements from 0 to 1.
a: 0
b: 1 + x0
B: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B.1(b.0(a.0(a.0(x)))) → B.1(b.0(x))
B.1(b.0(a.0(a.1(x)))) → B.0(b.1(x))
The TRS R consists of the following rules:
b.1(b.0(a.0(a.0(x)))) → a.1(b.0(b.1(b.0(x))))
b.1(b.0(a.0(a.1(x)))) → a.0(b.1(b.0(b.1(x))))
a.1(b.0(x)) → a.0(a.0(a.0(a.0(x))))
a.0(b.1(x)) → a.0(a.0(a.0(a.1(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B.1(b.0(a.0(a.0(x)))) → B.1(b.0(x))
The TRS R consists of the following rules:
b.1(b.0(a.0(a.0(x)))) → a.1(b.0(b.1(b.0(x))))
b.1(b.0(a.0(a.1(x)))) → a.0(b.1(b.0(b.1(x))))
a.1(b.0(x)) → a.0(a.0(a.0(a.0(x))))
a.0(b.1(x)) → a.0(a.0(a.0(a.1(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) 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:
B.1(b.0(a.0(a.0(x)))) → B.1(b.0(x))
The following rules are removed from R:
b.1(b.0(a.0(a.0(x)))) → a.1(b.0(b.1(b.0(x))))
b.1(b.0(a.0(a.1(x)))) → a.0(b.1(b.0(b.1(x))))
a.1(b.0(x)) → a.0(a.0(a.0(a.0(x))))
a.0(b.1(x)) → a.0(a.0(a.0(a.1(x))))
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(B.1(x1)) = x1
POL(a.0(x1)) = x1
POL(b.0(x1)) = x1
(20) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) YES