Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(x1))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(x1))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(x1))
The set Q is empty.
We have obtained the following QTRS:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
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:
B(d(x1)) → C(x1)
A(d(x1)) → A(x1)
A(c(b(x1))) → A(d(b(x1)))
A(d(x1)) → A(a(a(x1)))
C(a(a(x1))) → A(c(x1))
B(d(x1)) → B(c(x1))
C(a(a(x1))) → C(x1)
A(d(x1)) → A(a(x1))
The TRS R consists of the following rules:
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(d(x1)) → C(x1)
A(d(x1)) → A(x1)
A(c(b(x1))) → A(d(b(x1)))
A(d(x1)) → A(a(a(x1)))
C(a(a(x1))) → A(c(x1))
B(d(x1)) → B(c(x1))
C(a(a(x1))) → C(x1)
A(d(x1)) → A(a(x1))
The TRS R consists of the following rules:
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(d(x1)) → A(x1)
A(c(b(x1))) → A(d(b(x1)))
A(d(x1)) → A(a(a(x1)))
A(d(x1)) → A(a(x1))
The TRS R consists of the following rules:
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(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(d(x1)) → A(x1)
A(d(x1)) → A(a(a(x1)))
A(d(x1)) → A(a(x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = x1
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + 2·x1
POL(d(x1)) = 1 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(b(x1))) → A(d(b(x1)))
The TRS R consists of the following rules:
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(a(x1))) → C(x1)
The TRS R consists of the following rules:
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(a(x1))) → C(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(a(x1))) → C(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:
C(a(a(x1))) → C(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(C(x1)) = 2·x1
POL(a(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ 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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(d(x1)) → B(c(x1))
The TRS R consists of the following rules:
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(d(x1)) → B(c(x1)) at position [0] we obtained the following new rules:
B(d(a(a(x0)))) → B(a(c(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(d(a(a(x0)))) → B(a(c(x0)))
The TRS R consists of the following rules:
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(x1))
B(d(a(a(x0)))) → B(a(c(x0)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(x1))
B(d(a(a(x0)))) → B(a(c(x0)))
The set Q is empty.
We have obtained the following QTRS:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
a(a(d(B(x)))) → c(a(B(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
a(a(d(B(x)))) → c(a(B(x)))
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:
D(a(x)) → D(x)
B1(c(a(x))) → D(a(x))
B1(c(a(x))) → B1(d(a(x)))
D(a(x)) → A(d(x))
A(a(d(B(x)))) → A(B(x))
D(a(x)) → A(a(a(d(x))))
A(a(c(x))) → A(x)
D(a(x)) → A(a(d(x)))
The TRS R consists of the following rules:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
a(a(d(B(x)))) → c(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
D(a(x)) → D(x)
B1(c(a(x))) → D(a(x))
B1(c(a(x))) → B1(d(a(x)))
D(a(x)) → A(d(x))
A(a(d(B(x)))) → A(B(x))
D(a(x)) → A(a(a(d(x))))
A(a(c(x))) → A(x)
D(a(x)) → A(a(d(x)))
The TRS R consists of the following rules:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
a(a(d(B(x)))) → c(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(c(x))) → A(x)
The TRS R consists of the following rules:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
a(a(d(B(x)))) → c(a(B(x)))
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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(c(x))) → A(x)
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:
A(a(c(x))) → A(x)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(a(x1)) = 2·x1
POL(c(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ 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.
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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(c(x))) → A(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
D(a(x)) → D(x)
The TRS R consists of the following rules:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
a(a(d(B(x)))) → c(a(B(x)))
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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
D(a(x)) → D(x)
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:
D(a(x)) → D(x)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(D(x1)) = 2·x1
POL(a(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QDP
↳ QTRS Reverse
↳ 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.
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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
D(a(x)) → D(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(c(a(x))) → B1(d(a(x)))
The TRS R consists of the following rules:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
a(a(d(B(x)))) → c(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(c(a(x))) → B1(d(a(x))) at position [0] we obtained the following new rules:
B1(c(a(a(d(B(x0)))))) → B1(d(c(a(B(x0)))))
B1(c(a(x0))) → B1(a(a(a(d(x0)))))
B1(c(a(a(c(x0))))) → B1(d(c(a(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(c(a(a(d(B(x0)))))) → B1(d(c(a(B(x0)))))
B1(c(a(a(c(x0))))) → B1(d(c(a(x0))))
B1(c(a(x0))) → B1(a(a(a(d(x0)))))
The TRS R consists of the following rules:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
a(a(d(B(x)))) → c(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(c(a(x0))) → B1(a(a(a(d(x0)))))
The TRS R consists of the following rules:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
a(a(d(B(x)))) → c(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
a(a(d(B(x)))) → c(a(B(x)))
The set Q is empty.
We have obtained the following QTRS:
c(a(a(x))) → a(c(x))
a(c(b(x))) → a(d(b(x)))
a(d(x)) → d(a(a(a(x))))
b(d(x)) → b(c(x))
B(d(a(a(x)))) → B(a(c(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(x))) → a(c(x))
a(c(b(x))) → a(d(b(x)))
a(d(x)) → d(a(a(a(x))))
b(d(x)) → b(c(x))
B(d(a(a(x)))) → B(a(c(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
a(a(d(B(x)))) → c(a(B(x)))
The set Q is empty.
We have obtained the following QTRS:
c(a(a(x))) → a(c(x))
a(c(b(x))) → a(d(b(x)))
a(d(x)) → d(a(a(a(x))))
b(d(x)) → b(c(x))
B(d(a(a(x)))) → B(a(c(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(x))) → a(c(x))
a(c(b(x))) → a(d(b(x)))
a(d(x)) → d(a(a(a(x))))
b(d(x)) → b(c(x))
B(d(a(a(x)))) → B(a(c(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
c(a(a(x1))) → a(c(x1))
a(c(b(x1))) → a(d(b(x1)))
a(d(x1)) → d(a(a(a(x1))))
b(d(x1)) → b(c(x1))
The set Q is empty.
We have obtained the following QTRS:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(c(x))) → c(a(x))
b(c(a(x))) → b(d(a(x)))
d(a(x)) → a(a(a(d(x))))
d(b(x)) → c(b(x))
Q is empty.