Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
R2(a(x1)) → A(a(a(r2(x1))))
B(l2(x1)) → R1(x1)
A(l1(x1)) → A(a(x1))
R2(b(x1)) → A(b(x1))
R2(a(x1)) → R2(x1)
A(a(l2(x1))) → A(x1)
B(l1(x1)) → R2(x1)
A(l1(x1)) → A(x1)
R2(a(x1)) → A(r2(x1))
R2(a(x1)) → A(a(r2(x1)))
R1(a(x1)) → A(a(a(r1(x1))))
R1(a(x1)) → A(r1(x1))
R1(a(x1)) → A(a(r1(x1)))
B(l2(x1)) → B(r1(x1))
B(l1(x1)) → B(r2(x1))
R1(a(x1)) → R1(x1)
A(l1(x1)) → A(a(a(x1)))
A(a(l2(x1))) → A(a(x1))
The TRS R consists of the following rules:
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
R2(a(x1)) → A(a(a(r2(x1))))
B(l2(x1)) → R1(x1)
A(l1(x1)) → A(a(x1))
R2(b(x1)) → A(b(x1))
R2(a(x1)) → R2(x1)
A(a(l2(x1))) → A(x1)
B(l1(x1)) → R2(x1)
A(l1(x1)) → A(x1)
R2(a(x1)) → A(r2(x1))
R2(a(x1)) → A(a(r2(x1)))
R1(a(x1)) → A(a(a(r1(x1))))
R1(a(x1)) → A(r1(x1))
R1(a(x1)) → A(a(r1(x1)))
B(l2(x1)) → B(r1(x1))
B(l1(x1)) → B(r2(x1))
R1(a(x1)) → R1(x1)
A(l1(x1)) → A(a(a(x1)))
A(a(l2(x1))) → A(a(x1))
The TRS R consists of the following rules:
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 9 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(l1(x1)) → A(a(x1))
A(a(l2(x1))) → A(x1)
A(l1(x1)) → A(a(a(x1)))
A(a(l2(x1))) → A(a(x1))
A(l1(x1)) → A(x1)
The TRS R consists of the following rules:
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(l1(x1)) → A(a(x1))
A(a(l2(x1))) → A(x1)
A(l1(x1)) → A(a(a(x1)))
A(l1(x1)) → A(x1)
A(a(l2(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
A(l1(x1)) → A(a(x1))
A(l1(x1)) → A(a(a(x1)))
A(l1(x1)) → A(x1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(a(x1)) = x1
POL(l1(x1)) = 2 + 2·x1
POL(l2(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(l2(x1))) → A(x1)
A(a(l2(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
A(a(l2(x1))) → A(x1)
A(a(l2(x1))) → A(a(x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = x1
POL(a(x1)) = x1
POL(l1(x1)) = 2·x1
POL(l2(x1)) = 1 + 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(l1(x1)) → A(a(x1))
A(a(l2(x1))) → A(x1)
A(l1(x1)) → A(a(a(x1)))
A(l1(x1)) → A(x1)
A(a(l2(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
R2(a(x1)) → R2(x1)
The TRS R consists of the following rules:
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
R2(a(x1)) → R2(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
R2(a(x1)) → R2(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
R2(a(x1)) → R2(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(R2(x1)) = 2·x1
POL(a(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
R1(a(x1)) → R1(x1)
The TRS R consists of the following rules:
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
R1(a(x1)) → R1(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
R1(a(x1)) → R1(x1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
R1(a(x1)) → R1(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(R1(x1)) = 2·x1
POL(a(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(l1(x1)) → B(r2(x1))
B(l2(x1)) → B(r1(x1))
The TRS R consists of the following rules:
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(l2(x1)) → B(r1(x1)) at position [0] we obtained the following new rules:
B(l2(b(x0))) → B(l1(b(x0)))
B(l2(a(x0))) → B(a(a(a(r1(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(l2(b(x0))) → B(l1(b(x0)))
B(l1(x1)) → B(r2(x1))
B(l2(a(x0))) → B(a(a(a(r1(x0)))))
The TRS R consists of the following rules:
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(l1(x1)) → B(r2(x1)) at position [0] we obtained the following new rules:
B(l1(b(x0))) → B(l2(a(b(x0))))
B(l1(a(x0))) → B(a(a(a(r2(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(l1(a(x0))) → B(a(a(a(r2(x0)))))
B(l2(b(x0))) → B(l1(b(x0)))
B(l1(b(x0))) → B(l2(a(b(x0))))
B(l2(a(x0))) → B(a(a(a(r1(x0)))))
The TRS R consists of the following rules:
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.l2: 0
B: 0
a: 1 + x0
b: 0
l1: x0
r1: x0
r2: x0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
B.0(l1.0(b.1(x0))) → B.0(l2.1(a.0(b.1(x0))))
B.0(l2.0(b.0(x0))) → B.0(l1.0(b.0(x0)))
B.0(l2.1(a.0(x0))) → B.1(a.0(a.1(a.0(r1.0(x0)))))
B.0(l2.0(b.1(x0))) → B.0(l1.0(b.1(x0)))
B.0(l2.0(a.1(x0))) → B.0(a.1(a.0(a.1(r1.1(x0)))))
B.0(l1.0(a.1(x0))) → B.0(a.1(a.0(a.1(r2.1(x0)))))
B.1(l1.1(a.0(x0))) → B.1(a.0(a.1(a.0(r2.0(x0)))))
B.0(l1.0(b.0(x0))) → B.0(l2.1(a.0(b.0(x0))))
The TRS R consists of the following rules:
r2.0(a.1(x1)) → a.1(a.0(a.1(r2.1(x1))))
r1.1(a.0(x1)) → a.0(a.1(a.0(r1.0(x1))))
a.1(a.0(l2.0(x1))) → l2.0(a.1(a.0(x1)))
r1.0(b.1(x1)) → l1.0(b.1(x1))
b.1(l1.1(x1)) → b.1(r2.1(x1))
a.1(a.0(l2.1(x1))) → l2.1(a.0(a.1(x1)))
r2.0(b.0(x1)) → l2.1(a.0(b.0(x1)))
r1.0(a.1(x1)) → a.1(a.0(a.1(r1.1(x1))))
a.0(l1.0(x1)) → l1.1(a.0(a.1(a.0(x1))))
a.1(a.0(x1)) → x1
r2.0(b.1(x1)) → l2.1(a.0(b.1(x1)))
a.0(a.1(x1)) → x1
b.0(l2.0(x1)) → b.0(r1.0(x1))
r2.1(a.0(x1)) → a.0(a.1(a.0(r2.0(x1))))
r1.0(b.0(x1)) → l1.0(b.0(x1))
b.0(l2.1(x1)) → b.1(r1.1(x1))
b.0(l1.0(x1)) → b.0(r2.0(x1))
a.1(l1.1(x1)) → l1.0(a.1(a.0(a.1(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B.0(l1.0(b.1(x0))) → B.0(l2.1(a.0(b.1(x0))))
B.0(l2.0(b.0(x0))) → B.0(l1.0(b.0(x0)))
B.0(l2.1(a.0(x0))) → B.1(a.0(a.1(a.0(r1.0(x0)))))
B.0(l2.0(b.1(x0))) → B.0(l1.0(b.1(x0)))
B.0(l2.0(a.1(x0))) → B.0(a.1(a.0(a.1(r1.1(x0)))))
B.0(l1.0(a.1(x0))) → B.0(a.1(a.0(a.1(r2.1(x0)))))
B.1(l1.1(a.0(x0))) → B.1(a.0(a.1(a.0(r2.0(x0)))))
B.0(l1.0(b.0(x0))) → B.0(l2.1(a.0(b.0(x0))))
The TRS R consists of the following rules:
r2.0(a.1(x1)) → a.1(a.0(a.1(r2.1(x1))))
r1.1(a.0(x1)) → a.0(a.1(a.0(r1.0(x1))))
a.1(a.0(l2.0(x1))) → l2.0(a.1(a.0(x1)))
r1.0(b.1(x1)) → l1.0(b.1(x1))
b.1(l1.1(x1)) → b.1(r2.1(x1))
a.1(a.0(l2.1(x1))) → l2.1(a.0(a.1(x1)))
r2.0(b.0(x1)) → l2.1(a.0(b.0(x1)))
r1.0(a.1(x1)) → a.1(a.0(a.1(r1.1(x1))))
a.0(l1.0(x1)) → l1.1(a.0(a.1(a.0(x1))))
a.1(a.0(x1)) → x1
r2.0(b.1(x1)) → l2.1(a.0(b.1(x1)))
a.0(a.1(x1)) → x1
b.0(l2.0(x1)) → b.0(r1.0(x1))
r2.1(a.0(x1)) → a.0(a.1(a.0(r2.0(x1))))
r1.0(b.0(x1)) → l1.0(b.0(x1))
b.0(l2.1(x1)) → b.1(r1.1(x1))
b.0(l1.0(x1)) → b.0(r2.0(x1))
a.1(l1.1(x1)) → l1.0(a.1(a.0(a.1(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B.1(l1.1(a.0(x0))) → B.1(a.0(a.1(a.0(r2.0(x0)))))
The TRS R consists of the following rules:
r2.0(a.1(x1)) → a.1(a.0(a.1(r2.1(x1))))
r1.1(a.0(x1)) → a.0(a.1(a.0(r1.0(x1))))
a.1(a.0(l2.0(x1))) → l2.0(a.1(a.0(x1)))
r1.0(b.1(x1)) → l1.0(b.1(x1))
b.1(l1.1(x1)) → b.1(r2.1(x1))
a.1(a.0(l2.1(x1))) → l2.1(a.0(a.1(x1)))
r2.0(b.0(x1)) → l2.1(a.0(b.0(x1)))
r1.0(a.1(x1)) → a.1(a.0(a.1(r1.1(x1))))
a.0(l1.0(x1)) → l1.1(a.0(a.1(a.0(x1))))
a.1(a.0(x1)) → x1
r2.0(b.1(x1)) → l2.1(a.0(b.1(x1)))
a.0(a.1(x1)) → x1
b.0(l2.0(x1)) → b.0(r1.0(x1))
r2.1(a.0(x1)) → a.0(a.1(a.0(r2.0(x1))))
r1.0(b.0(x1)) → l1.0(b.0(x1))
b.0(l2.1(x1)) → b.1(r1.1(x1))
b.0(l1.0(x1)) → b.0(r2.0(x1))
a.1(l1.1(x1)) → l1.0(a.1(a.0(a.1(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
r1.0(b.1(x1)) → l1.0(b.1(x1))
r1.0(b.0(x1)) → l1.0(b.0(x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(B.1(x1)) = x1
POL(a.0(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = 1 + x1
POL(b.1(x1)) = x1
POL(l1.0(x1)) = x1
POL(l1.1(x1)) = x1
POL(l2.0(x1)) = 1 + x1
POL(l2.1(x1)) = x1
POL(r1.0(x1)) = 1 + x1
POL(r1.1(x1)) = 1 + x1
POL(r2.0(x1)) = x1
POL(r2.1(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B.1(l1.1(a.0(x0))) → B.1(a.0(a.1(a.0(r2.0(x0)))))
The TRS R consists of the following rules:
r2.0(a.1(x1)) → a.1(a.0(a.1(r2.1(x1))))
r1.1(a.0(x1)) → a.0(a.1(a.0(r1.0(x1))))
a.1(a.0(l2.0(x1))) → l2.0(a.1(a.0(x1)))
b.1(l1.1(x1)) → b.1(r2.1(x1))
a.1(a.0(l2.1(x1))) → l2.1(a.0(a.1(x1)))
r2.0(b.0(x1)) → l2.1(a.0(b.0(x1)))
r1.0(a.1(x1)) → a.1(a.0(a.1(r1.1(x1))))
a.0(l1.0(x1)) → l1.1(a.0(a.1(a.0(x1))))
a.1(a.0(x1)) → x1
r2.0(b.1(x1)) → l2.1(a.0(b.1(x1)))
a.0(a.1(x1)) → x1
b.0(l2.0(x1)) → b.0(r1.0(x1))
r2.1(a.0(x1)) → a.0(a.1(a.0(r2.0(x1))))
b.0(l2.1(x1)) → b.1(r1.1(x1))
b.0(l1.0(x1)) → b.0(r2.0(x1))
a.1(l1.1(x1)) → l1.0(a.1(a.0(a.1(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
B.1(l1.1(a.0(x0))) → B.1(a.0(a.1(a.0(r2.0(x0)))))
Strictly oriented rules of the TRS R:
b.1(l1.1(x1)) → b.1(r2.1(x1))
b.0(l2.0(x1)) → b.0(r1.0(x1))
b.0(l2.1(x1)) → b.1(r1.1(x1))
b.0(l1.0(x1)) → b.0(r2.0(x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(B.1(x1)) = x1
POL(a.0(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = 1 + x1
POL(b.1(x1)) = x1
POL(l1.0(x1)) = 1 + x1
POL(l1.1(x1)) = 1 + x1
POL(l2.0(x1)) = 1 + x1
POL(l2.1(x1)) = x1
POL(r1.0(x1)) = x1
POL(r1.1(x1)) = x1
POL(r2.0(x1)) = x1
POL(r2.1(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
The TRS R consists of the following rules:
r2.0(a.1(x1)) → a.1(a.0(a.1(r2.1(x1))))
r1.1(a.0(x1)) → a.0(a.1(a.0(r1.0(x1))))
a.1(a.0(l2.0(x1))) → l2.0(a.1(a.0(x1)))
a.1(a.0(l2.1(x1))) → l2.1(a.0(a.1(x1)))
r2.0(b.0(x1)) → l2.1(a.0(b.0(x1)))
r1.0(a.1(x1)) → a.1(a.0(a.1(r1.1(x1))))
a.0(l1.0(x1)) → l1.1(a.0(a.1(a.0(x1))))
a.1(a.0(x1)) → x1
r2.0(b.1(x1)) → l2.1(a.0(b.1(x1)))
a.0(a.1(x1)) → x1
r2.1(a.0(x1)) → a.0(a.1(a.0(r2.0(x1))))
a.1(l1.1(x1)) → l1.0(a.1(a.0(a.1(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B.0(l2.0(b.0(x0))) → B.0(l1.0(b.0(x0)))
B.0(l2.0(b.1(x0))) → B.0(l1.0(b.1(x0)))
B.0(l2.0(a.1(x0))) → B.0(a.1(a.0(a.1(r1.1(x0)))))
B.0(l1.0(a.1(x0))) → B.0(a.1(a.0(a.1(r2.1(x0)))))
The TRS R consists of the following rules:
r2.0(a.1(x1)) → a.1(a.0(a.1(r2.1(x1))))
r1.1(a.0(x1)) → a.0(a.1(a.0(r1.0(x1))))
a.1(a.0(l2.0(x1))) → l2.0(a.1(a.0(x1)))
r1.0(b.1(x1)) → l1.0(b.1(x1))
b.1(l1.1(x1)) → b.1(r2.1(x1))
a.1(a.0(l2.1(x1))) → l2.1(a.0(a.1(x1)))
r2.0(b.0(x1)) → l2.1(a.0(b.0(x1)))
r1.0(a.1(x1)) → a.1(a.0(a.1(r1.1(x1))))
a.0(l1.0(x1)) → l1.1(a.0(a.1(a.0(x1))))
a.1(a.0(x1)) → x1
r2.0(b.1(x1)) → l2.1(a.0(b.1(x1)))
a.0(a.1(x1)) → x1
b.0(l2.0(x1)) → b.0(r1.0(x1))
r2.1(a.0(x1)) → a.0(a.1(a.0(r2.0(x1))))
r1.0(b.0(x1)) → l1.0(b.0(x1))
b.0(l2.1(x1)) → b.1(r1.1(x1))
b.0(l1.0(x1)) → b.0(r2.0(x1))
a.1(l1.1(x1)) → l1.0(a.1(a.0(a.1(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
B.0(l2.0(b.0(x0))) → B.0(l1.0(b.0(x0)))
B.0(l2.0(b.1(x0))) → B.0(l1.0(b.1(x0)))
Strictly oriented rules of the TRS R:
r1.0(b.1(x1)) → l1.0(b.1(x1))
r1.0(b.0(x1)) → l1.0(b.0(x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(B.0(x1)) = x1
POL(a.0(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = 1 + x1
POL(b.1(x1)) = x1
POL(l1.0(x1)) = x1
POL(l1.1(x1)) = x1
POL(l2.0(x1)) = 1 + x1
POL(l2.1(x1)) = x1
POL(r1.0(x1)) = 1 + x1
POL(r1.1(x1)) = 1 + x1
POL(r2.0(x1)) = x1
POL(r2.1(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B.0(l1.0(a.1(x0))) → B.0(a.1(a.0(a.1(r2.1(x0)))))
B.0(l2.0(a.1(x0))) → B.0(a.1(a.0(a.1(r1.1(x0)))))
The TRS R consists of the following rules:
r2.0(a.1(x1)) → a.1(a.0(a.1(r2.1(x1))))
r1.1(a.0(x1)) → a.0(a.1(a.0(r1.0(x1))))
a.1(a.0(l2.0(x1))) → l2.0(a.1(a.0(x1)))
b.1(l1.1(x1)) → b.1(r2.1(x1))
a.1(a.0(l2.1(x1))) → l2.1(a.0(a.1(x1)))
r2.0(b.0(x1)) → l2.1(a.0(b.0(x1)))
r1.0(a.1(x1)) → a.1(a.0(a.1(r1.1(x1))))
a.0(l1.0(x1)) → l1.1(a.0(a.1(a.0(x1))))
a.1(a.0(x1)) → x1
r2.0(b.1(x1)) → l2.1(a.0(b.1(x1)))
a.0(a.1(x1)) → x1
b.0(l2.0(x1)) → b.0(r1.0(x1))
r2.1(a.0(x1)) → a.0(a.1(a.0(r2.0(x1))))
b.0(l2.1(x1)) → b.1(r1.1(x1))
b.0(l1.0(x1)) → b.0(r2.0(x1))
a.1(l1.1(x1)) → l1.0(a.1(a.0(a.1(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
B.0(l1.0(a.1(x0))) → B.0(a.1(a.0(a.1(r2.1(x0)))))
B.0(l2.0(a.1(x0))) → B.0(a.1(a.0(a.1(r1.1(x0)))))
Strictly oriented rules of the TRS R:
b.1(l1.1(x1)) → b.1(r2.1(x1))
b.0(l2.0(x1)) → b.0(r1.0(x1))
b.0(l2.1(x1)) → b.1(r1.1(x1))
b.0(l1.0(x1)) → b.0(r2.0(x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(B.0(x1)) = x1
POL(a.0(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = 1 + x1
POL(b.1(x1)) = x1
POL(l1.0(x1)) = 1 + x1
POL(l1.1(x1)) = 1 + x1
POL(l2.0(x1)) = 1 + x1
POL(l2.1(x1)) = x1
POL(r1.0(x1)) = x1
POL(r1.1(x1)) = x1
POL(r2.0(x1)) = x1
POL(r2.1(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
The TRS R consists of the following rules:
r2.0(a.1(x1)) → a.1(a.0(a.1(r2.1(x1))))
r1.1(a.0(x1)) → a.0(a.1(a.0(r1.0(x1))))
a.1(a.0(l2.0(x1))) → l2.0(a.1(a.0(x1)))
a.1(a.0(l2.1(x1))) → l2.1(a.0(a.1(x1)))
r2.0(b.0(x1)) → l2.1(a.0(b.0(x1)))
r1.0(a.1(x1)) → a.1(a.0(a.1(r1.1(x1))))
a.0(l1.0(x1)) → l1.1(a.0(a.1(a.0(x1))))
a.1(a.0(x1)) → x1
r2.0(b.1(x1)) → l2.1(a.0(b.1(x1)))
a.0(a.1(x1)) → x1
r2.1(a.0(x1)) → a.0(a.1(a.0(r2.0(x1))))
a.1(l1.1(x1)) → l1.0(a.1(a.0(a.1(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We have reversed the following QTRS:
The set of rules R is
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(r1(x)) → r1(a(a(a(x))))
a(r2(x)) → r2(a(a(a(x))))
l1(a(x)) → a(a(a(l1(x))))
l2(a(a(x))) → a(a(l2(x)))
b(r1(x)) → b(l1(x))
b(r2(x)) → b(a(l2(x)))
l1(b(x)) → r2(b(x))
l2(b(x)) → r1(b(x))
a(a(x)) → x
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(r1(x)) → r1(a(a(a(x))))
a(r2(x)) → r2(a(a(a(x))))
l1(a(x)) → a(a(a(l1(x))))
l2(a(a(x))) → a(a(l2(x)))
b(r1(x)) → b(l1(x))
b(r2(x)) → b(a(l2(x)))
l1(b(x)) → r2(b(x))
l2(b(x)) → r1(b(x))
a(a(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
r1(a(x1)) → a(a(a(r1(x1))))
r2(a(x1)) → a(a(a(r2(x1))))
a(l1(x1)) → l1(a(a(a(x1))))
a(a(l2(x1))) → l2(a(a(x1)))
r1(b(x1)) → l1(b(x1))
r2(b(x1)) → l2(a(b(x1)))
b(l1(x1)) → b(r2(x1))
b(l2(x1)) → b(r1(x1))
a(a(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(r1(x)) → r1(a(a(a(x))))
a(r2(x)) → r2(a(a(a(x))))
l1(a(x)) → a(a(a(l1(x))))
l2(a(a(x))) → a(a(l2(x)))
b(r1(x)) → b(l1(x))
b(r2(x)) → b(a(l2(x)))
l1(b(x)) → r2(b(x))
l2(b(x)) → r1(b(x))
a(a(x)) → x
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(r1(x)) → r1(a(a(a(x))))
a(r2(x)) → r2(a(a(a(x))))
l1(a(x)) → a(a(a(l1(x))))
l2(a(a(x))) → a(a(l2(x)))
b(r1(x)) → b(l1(x))
b(r2(x)) → b(a(l2(x)))
l1(b(x)) → r2(b(x))
l2(b(x)) → r1(b(x))
a(a(x)) → x
Q is empty.