(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(x)))) → b(b(b(a(a(a(x))))))
a(c(x)) → c(a(x))
b(c(x)) → c(b(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(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
c(b(x)) → b(c(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)))) → B(b(b(x)))
B(b(a(a(x)))) → B(b(x))
B(b(a(a(x)))) → B(x)
C(a(x)) → C(x)
C(b(x)) → B(c(x))
C(b(x)) → C(x)
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
c(b(x)) → b(c(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 1 less node.
(6) Complex Obligation (AND)
(7) 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(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
c(b(x)) → b(c(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:
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(a(a(b(b(b(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(11) 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(a(a(b(b(b(x))))))
The set Q consists of the following terms:
b(b(a(a(x0))))
We have to consider all minimal (P,Q,R)-chains.
(12) 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 natural numbers [ARCTIC]:
| POL(b(x1)) = | | + | | / | -I | -I | 0A | \ |
| | | 0A | -I | 1A | | |
| \ | 0A | -I | -I | / |
| · | x1 |
| POL(a(x1)) = | | + | | / | -I | 0A | 0A | \ |
| | | 0A | -I | 1A | | |
| \ | -I | -I | -I | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
(13) 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(a(a(b(b(b(x))))))
The set Q consists of the following terms:
b(b(a(a(x0))))
We have to consider all minimal (P,Q,R)-chains.
(14) 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(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^5, +, *, >=, >) :
| POL(b(x1)) = | | + | | / | 0 | 1 | 0 | 0 | 0 | \ |
| | | 0 | 0 | 0 | 0 | 1 | | |
| | | 1 | 0 | 0 | 0 | 1 | | |
| | | 0 | 0 | 0 | 0 | 0 | | |
| \ | 0 | 1 | 0 | 0 | 0 | / |
| · | x1 |
| POL(a(x1)) = | | + | | / | 0 | 0 | 0 | 0 | 0 | \ |
| | | 0 | 0 | 1 | 1 | 0 | | |
| | | 0 | 1 | 0 | 0 | 1 | | |
| | | 0 | 0 | 0 | 0 | 1 | | |
| \ | 0 | 0 | 0 | 0 | 0 | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
(15) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
The set Q consists of the following terms:
b(b(a(a(x0))))
We have to consider all minimal (P,Q,R)-chains.
(16) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(17) YES
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → C(x)
C(a(x)) → C(x)
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(b(b(b(x))))))
c(a(x)) → a(c(x))
c(b(x)) → b(c(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) 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.
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → C(x)
C(a(x)) → C(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) 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:
- C(b(x)) → C(x)
The graph contains the following edges 1 > 1
- C(a(x)) → C(x)
The graph contains the following edges 1 > 1
(22) YES