(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(b(b(b(x))))
b(a(x)) → a(a(a(a(x))))
a(x) → x
b(x) → 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(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → 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(x)) → B(b(b(b(x))))
B(a(x)) → B(b(b(x)))
B(a(x)) → B(b(x))
B(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(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → 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.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(a(a(x)))
A(b(x)) → A(a(a(a(x))))
A(b(x)) → A(a(x))
A(b(x)) → A(x)
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → 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(a(a(x)))
A(b(x)) → A(a(a(a(x))))
A(b(x)) → A(a(x))
A(b(x)) → A(x)
The TRS R consists of the following rules:
a(b(x)) → a(a(a(a(x))))
a(x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) 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:
A(b(x)) → A(a(a(x)))
A(b(x)) → A(a(a(a(x))))
A(b(x)) → A(a(x))
A(b(x)) → A(x)
Strictly oriented rules of the TRS R:
a(b(x)) → a(a(a(a(x))))
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = x1
POL(a(x1)) = x1
POL(b(x1)) = 1 + x1
(11) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(x)) → B(b(b(x)))
B(a(x)) → B(b(b(b(x))))
B(a(x)) → B(b(x))
B(a(x)) → B(x)
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(x)) → B(b(b(x)))
B(a(x)) → B(b(b(b(x))))
B(a(x)) → B(b(x))
B(a(x)) → B(x)
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
b(x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(a(x)) → B(b(b(x)))
B(a(x)) → B(b(b(b(x))))
B(a(x)) → B(b(x))
B(a(x)) → B(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( B(x1) ) = max{0, 2x1 - 2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
b(a(x)) → b(b(b(b(x))))
b(x) → x
(18) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
b(x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(20) YES