(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(s(x)) → s(a(x))
b(a(b(s(x)))) → a(b(s(a(x))))
b(a(b(b(x)))) → a(b(a(b(x))))
a(b(a(a(x)))) → b(a(b(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:
s(a(x)) → a(s(x))
s(b(a(b(x)))) → a(s(b(a(x))))
b(b(a(b(x)))) → b(a(b(a(x))))
a(a(b(a(x)))) → a(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:
S(a(x)) → A(s(x))
S(a(x)) → S(x)
S(b(a(b(x)))) → A(s(b(a(x))))
S(b(a(b(x)))) → S(b(a(x)))
S(b(a(b(x)))) → B(a(x))
S(b(a(b(x)))) → A(x)
B(b(a(b(x)))) → B(a(b(a(x))))
B(b(a(b(x)))) → A(b(a(x)))
B(b(a(b(x)))) → B(a(x))
B(b(a(b(x)))) → A(x)
A(a(b(a(x)))) → A(b(a(b(x))))
A(a(b(a(x)))) → B(a(b(x)))
A(a(b(a(x)))) → A(b(x))
A(a(b(a(x)))) → B(x)
The TRS R consists of the following rules:
s(a(x)) → a(s(x))
s(b(a(b(x)))) → a(s(b(a(x))))
b(b(a(b(x)))) → b(a(b(a(x))))
a(a(b(a(x)))) → a(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 2 SCCs with 4 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(a(b(x)))) → A(b(a(x)))
A(a(b(a(x)))) → A(b(a(b(x))))
A(a(b(a(x)))) → B(a(b(x)))
B(b(a(b(x)))) → B(a(b(a(x))))
B(b(a(b(x)))) → B(a(x))
B(b(a(b(x)))) → A(x)
A(a(b(a(x)))) → A(b(x))
A(a(b(a(x)))) → B(x)
The TRS R consists of the following rules:
s(a(x)) → a(s(x))
s(b(a(b(x)))) → a(s(b(a(x))))
b(b(a(b(x)))) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(b(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(b(x)))) → A(b(a(x)))
A(a(b(a(x)))) → A(b(a(b(x))))
A(a(b(a(x)))) → B(a(b(x)))
B(b(a(b(x)))) → B(a(b(a(x))))
B(b(a(b(x)))) → B(a(x))
B(b(a(b(x)))) → A(x)
A(a(b(a(x)))) → A(b(x))
A(a(b(a(x)))) → B(x)
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(b(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:
B(b(a(b(x)))) → A(b(a(x)))
A(a(b(a(x)))) → B(a(b(x)))
B(b(a(b(x)))) → B(a(x))
B(b(a(b(x)))) → A(x)
A(a(b(a(x)))) → A(b(x))
A(a(b(a(x)))) → B(x)
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 1 + 2·x1
POL(B(x1)) = 2 + 2·x1
POL(a(x1)) = 2 + x1
POL(b(x1)) = 2 + x1
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(a(b(a(x)))) → A(b(a(b(x))))
B(b(a(b(x)))) → B(a(b(a(x))))
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(b(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(13) Complex Obligation (AND)
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(a(b(x)))) → B(a(b(a(x))))
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(b(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(b(a(b(x)))) → B(a(b(a(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(B(x1)) = x1
POL(a(x1)) = 0
POL(b(x1)) = 1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(a(b(a(x)))) → a(b(a(b(x))))
(16) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(b(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(18) YES
(19) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(a(b(a(x)))) → A(b(a(b(x))))
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(b(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(20) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
A(a(b(a(x)))) → A(b(a(b(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = x1
POL(a(x1)) = 1
POL(b(x1)) = 0
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
b(b(a(b(x)))) → b(a(b(a(x))))
(21) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(b(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(22) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(23) YES
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(b(a(b(x)))) → S(b(a(x)))
S(a(x)) → S(x)
The TRS R consists of the following rules:
s(a(x)) → a(s(x))
s(b(a(b(x)))) → a(s(b(a(x))))
b(b(a(b(x)))) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(b(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) 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.
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(b(a(b(x)))) → S(b(a(x)))
S(a(x)) → S(x)
The TRS R consists of the following rules:
a(a(b(a(x)))) → a(b(a(b(x))))
b(b(a(b(x)))) → b(a(b(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(27) 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:
S(b(a(b(x)))) → S(b(a(x)))
S(a(x)) → S(x)
Used ordering: Polynomial interpretation [POLO]:
POL(S(x1)) = x1
POL(a(x1)) = 1 + x1
POL(b(x1)) = 1 + x1
(28) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(a(b(a(x)))) → a(b(a(b(x))))
b(b(a(b(x)))) → b(a(b(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(29) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(30) YES