(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(b(b(x))))) → b(b(b(x)))
b(a(a(a(b(x))))) → a(a(a(b(a(a(a(x)))))))
a(a(a(x))) → a(a(x))
b(b(x)) → a(b(a(x)))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(a(x1)) = 1 + x1
POL(b(x1)) = 3 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
a(a(a(x))) → a(a(x))
b(b(x)) → a(b(a(x)))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(b(b(x))))) → b(b(b(x)))
b(a(a(a(b(x))))) → a(a(a(b(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:
A(a(a(b(b(x))))) → B(b(b(x)))
B(a(a(a(b(x))))) → A(a(a(b(a(a(a(x)))))))
B(a(a(a(b(x))))) → A(a(b(a(a(a(x))))))
B(a(a(a(b(x))))) → A(b(a(a(a(x)))))
B(a(a(a(b(x))))) → B(a(a(a(x))))
B(a(a(a(b(x))))) → A(a(a(x)))
B(a(a(a(b(x))))) → A(a(x))
B(a(a(a(b(x))))) → A(x)
The TRS R consists of the following rules:
a(a(a(b(b(x))))) → b(b(b(x)))
b(a(a(a(b(x))))) → a(a(a(b(a(a(a(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
B(a(a(a(b(x))))) → A(a(b(a(a(a(x))))))
B(a(a(a(b(x))))) → A(b(a(a(a(x)))))
B(a(a(a(b(x))))) → B(a(a(a(x))))
B(a(a(a(b(x))))) → A(a(a(x)))
B(a(a(a(b(x))))) → A(a(x))
B(a(a(a(b(x))))) → A(x)
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = x1
POL(B(x1)) = 2 + x1
POL(a(x1)) = 1 + x1
POL(b(x1)) = 3 + x1
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(b(x))))) → B(b(b(x)))
B(a(a(a(b(x))))) → A(a(a(b(a(a(a(x)))))))
The TRS R consists of the following rules:
a(a(a(b(b(x))))) → b(b(b(x)))
b(a(a(a(b(x))))) → a(a(a(b(a(a(a(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.
A(a(a(b(b(x))))) → B(b(b(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(a(x1)) = | | + | / | 0A | -I | 0A | \ |
| | 0A | -I | -I | | |
\ | 0A | -I | -I | / |
| · | x1 |
POL(b(x1)) = | | + | / | 0A | -I | -I | \ |
| | 1A | -I | -I | | |
\ | 0A | 0A | 0A | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
b(a(a(a(b(x))))) → a(a(a(b(a(a(a(x)))))))
a(a(a(b(b(x))))) → b(b(b(x)))
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(b(x))))) → A(a(a(b(a(a(a(x)))))))
The TRS R consists of the following rules:
a(a(a(b(b(x))))) → b(b(b(x)))
b(a(a(a(b(x))))) → a(a(a(b(a(a(a(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 0 SCCs with 1 less node.
(10) TRUE