(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → x
a(b(a(b(x)))) → b(a(b(b(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:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(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(a(b(a(x)))) → A(a(b(b(a(b(x))))))
B(a(b(a(x)))) → A(b(b(a(b(x)))))
B(a(b(a(x)))) → B(b(a(b(x))))
B(a(b(a(x)))) → B(a(b(x)))
B(a(b(a(x)))) → A(b(x))
B(a(b(a(x)))) → B(x)
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(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 1 SCC with 3 less nodes.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(b(a(x)))) → B(a(b(x)))
B(a(b(a(x)))) → B(b(a(b(x))))
B(a(b(a(x)))) → B(x)
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(a(b(a(x)))) → B(b(a(b(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(a(x1)) = | | + | | / | -I | 0A | 0A | \ |
| | | 0A | -I | 0A | | |
| \ | -I | -I | 0A | / |
| · | x1 |
| POL(b(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | -I | -I | 0A | | |
| \ | -I | -I | -I | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
a(a(x)) → x
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(b(a(x)))) → B(a(b(x)))
B(a(b(a(x)))) → B(x)
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(a(b(a(x)))) → B(a(b(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(a(x1)) = | | + | | / | -I | 0A | 0A | \ |
| | | 0A | -I | -I | | |
| \ | -I | 0A | 1A | / |
| · | x1 |
| POL(b(x1)) = | | + | | / | -I | 0A | 1A | \ |
| | | -I | -I | -I | | |
| \ | -I | 0A | -I | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
b(a(b(a(x)))) → a(a(b(b(a(b(x))))))
a(a(x)) → x
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(b(a(x)))) → B(x)
The TRS R consists of the following rules:
a(a(x)) → x
b(a(b(a(x)))) → a(a(b(b(a(b(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:
B(a(b(a(x)))) → B(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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:
- B(a(b(a(x)))) → B(x)
The graph contains the following edges 1 > 1
(14) YES