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:
a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(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:
D(x1) → A(x1)
R(a(x1)) → D(r(x1))
D(a(x1)) → D(x1)
R(a(x1)) → R(x1)
D(a(x1)) → A(d(x1))
R(x1) → D(x1)
A(b(x1)) → R(x1)
D(a(x1)) → A(a(d(x1)))
The TRS R consists of the following rules:
a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
D(x1) → A(x1)
R(a(x1)) → D(r(x1))
D(a(x1)) → D(x1)
R(a(x1)) → R(x1)
D(a(x1)) → A(d(x1))
R(x1) → D(x1)
A(b(x1)) → R(x1)
D(a(x1)) → A(a(d(x1)))
The TRS R consists of the following rules:
a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(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:
R(a(x1)) → D(r(x1))
R(x1) → D(x1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(D(x1)) = 2·x1
POL(R(x1)) = 2 + 2·x1
POL(a(x1)) = x1
POL(b(x1)) = 1 + x1
POL(d(x1)) = x1
POL(r(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
D(x1) → A(x1)
R(a(x1)) → R(x1)
D(a(x1)) → D(x1)
D(a(x1)) → A(d(x1))
A(b(x1)) → R(x1)
D(a(x1)) → A(a(d(x1)))
The TRS R consists of the following rules:
a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(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 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
R(a(x1)) → R(x1)
The TRS R consists of the following rules:
a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(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
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
R(a(x1)) → R(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:
R(a(x1)) → R(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(R(x1)) = 2·x1
POL(a(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
↳ QDP
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
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
D(a(x1)) → D(x1)
The TRS R consists of the following rules:
a(b(x1)) → b(r(x1))
r(a(x1)) → d(r(x1))
r(x1) → d(x1)
d(a(x1)) → a(a(d(x1)))
d(x1) → a(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
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
Q DP problem:
The TRS P consists of the following rules:
D(a(x1)) → D(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:
D(a(x1)) → D(x1)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(D(x1)) = 2·x1
POL(a(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
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.