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(a(b(b(x1)))) → b(b(c(c(a(a(x1))))))
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
a(a(c(c(x1)))) → c(c(a(a(b(b(x1))))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(c(c(a(a(x1))))))
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
a(a(c(c(x1)))) → c(c(a(a(b(b(x1))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(x1)))) → b(b(c(c(a(a(x1))))))
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
a(a(c(c(x1)))) → c(c(a(a(b(b(x1))))))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))
c(c(b(b(x)))) → b(b(b(b(c(c(x))))))
c(c(a(a(x)))) → b(b(a(a(c(c(x))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))
c(c(b(b(x)))) → b(b(b(b(c(c(x))))))
c(c(a(a(x)))) → b(b(a(a(c(c(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(x1)))) → b(b(c(c(a(a(x1))))))
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
a(a(c(c(x1)))) → c(c(a(a(b(b(x1))))))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))
c(c(b(b(x)))) → b(b(b(b(c(c(x))))))
c(c(a(a(x)))) → b(b(a(a(c(c(x))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))
c(c(b(b(x)))) → b(b(b(b(c(c(x))))))
c(c(a(a(x)))) → b(b(a(a(c(c(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:
A(a(b(b(x1)))) → B(c(c(a(a(x1)))))
B(b(c(c(x1)))) → B(b(b(x1)))
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → B(b(c(c(a(a(x1))))))
A(a(c(c(x1)))) → B(x1)
A(a(c(c(x1)))) → B(b(x1))
A(a(c(c(x1)))) → A(b(b(x1)))
B(b(c(c(x1)))) → B(b(x1))
B(b(c(c(x1)))) → B(b(b(b(x1))))
A(a(b(b(x1)))) → A(x1)
A(a(c(c(x1)))) → A(a(b(b(x1))))
B(b(c(c(x1)))) → B(x1)
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(c(c(a(a(x1))))))
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
a(a(c(c(x1)))) → c(c(a(a(b(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A(a(b(b(x1)))) → B(c(c(a(a(x1)))))
B(b(c(c(x1)))) → B(b(b(x1)))
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → B(b(c(c(a(a(x1))))))
A(a(c(c(x1)))) → B(x1)
A(a(c(c(x1)))) → B(b(x1))
A(a(c(c(x1)))) → A(b(b(x1)))
B(b(c(c(x1)))) → B(b(x1))
B(b(c(c(x1)))) → B(b(b(b(x1))))
A(a(b(b(x1)))) → A(x1)
A(a(c(c(x1)))) → A(a(b(b(x1))))
B(b(c(c(x1)))) → B(x1)
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(c(c(a(a(x1))))))
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
a(a(c(c(x1)))) → c(c(a(a(b(b(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 5 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(b(c(c(x1)))) → B(b(b(x1)))
B(b(c(c(x1)))) → B(b(x1))
B(b(c(c(x1)))) → B(b(b(b(x1))))
B(b(c(c(x1)))) → B(x1)
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(c(c(a(a(x1))))))
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
a(a(c(c(x1)))) → c(c(a(a(b(b(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(b(c(c(x1)))) → B(b(b(x1)))
B(b(c(c(x1)))) → B(b(x1))
B(b(c(c(x1)))) → B(b(b(b(x1))))
B(b(c(c(x1)))) → B(x1)
The TRS R consists of the following rules:
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ RuleRemovalProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(b(c(c(x1)))) → B(b(b(x1)))
B(b(c(c(x1)))) → B(b(x1))
B(b(c(c(x1)))) → B(b(b(b(x1))))
B(b(c(c(x1)))) → B(x1)
The TRS R consists of the following rules:
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
The set Q consists of the following terms:
b(b(c(c(x0))))
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(b(c(c(x1)))) → B(b(b(x1)))
B(b(c(c(x1)))) → B(b(x1))
B(b(c(c(x1)))) → B(b(b(b(x1))))
B(b(c(c(x1)))) → B(x1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(B(x1)) = 2·x1
POL(b(x1)) = x1
POL(c(x1)) = 2 + x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ UsableRulesProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
The set Q consists of the following terms:
b(b(c(c(x0))))
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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(b(c(c(x1)))) → B(b(b(x1)))
B(b(c(c(x1)))) → B(b(x1))
B(b(c(c(x1)))) → B(b(b(b(x1))))
B(b(c(c(x1)))) → B(x1)
The TRS R consists of the following rules:
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(b(c(c(x1)))) → B(b(b(x1)))
B(b(c(c(x1)))) → B(b(x1))
B(b(c(c(x1)))) → B(b(b(b(x1))))
B(b(c(c(x1)))) → B(x1)
The TRS R consists of the following rules:
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
The set Q consists of the following terms:
b(b(c(c(x0))))
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(b(c(c(x1)))) → B(b(b(x1)))
B(b(c(c(x1)))) → B(b(x1))
B(b(c(c(x1)))) → B(b(b(b(x1))))
B(b(c(c(x1)))) → B(x1)
The TRS R consists of the following rules:
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
Q is empty.
We have to consider all (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(x1)
A(a(c(c(x1)))) → A(a(b(b(x1))))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(c(c(a(a(x1))))))
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
a(a(c(c(x1)))) → c(c(a(a(b(b(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(b(b(x1)))) → A(x1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = x1
POL(a(x1)) = 1 + 2·x1
POL(b(x1)) = x1
POL(c(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ SemLabProof
Q DP problem:
The TRS P consists of the following rules:
A(a(b(b(x1)))) → A(a(x1))
A(a(c(c(x1)))) → A(a(b(b(x1))))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(c(c(a(a(x1))))))
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
a(a(c(c(x1)))) → c(c(a(a(b(b(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.c: 0
a: 1 + x0
A: 0
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
A.1(a.0(c.0(c.0(x1)))) → A.1(a.0(b.0(b.0(x1))))
A.1(a.0(b.0(b.0(x1)))) → A.1(a.0(x1))
A.1(a.0(b.0(b.1(x1)))) → A.0(a.1(x1))
A.1(a.0(c.0(c.1(x1)))) → A.1(a.0(b.0(b.1(x1))))
The TRS R consists of the following rules:
a.1(a.0(c.0(c.0(x1)))) → c.0(c.0(a.1(a.0(b.0(b.0(x1))))))
b.0(b.0(c.0(c.1(x1)))) → c.0(c.0(b.0(b.0(b.0(b.1(x1))))))
b.0(b.0(c.0(c.0(x1)))) → c.0(c.0(b.0(b.0(b.0(b.0(x1))))))
a.1(a.0(b.0(b.1(x1)))) → b.0(b.0(c.0(c.1(a.0(a.1(x1))))))
a.1(a.0(b.0(b.0(x1)))) → b.0(b.0(c.0(c.0(a.1(a.0(x1))))))
a.1(a.0(c.0(c.1(x1)))) → c.0(c.0(a.1(a.0(b.0(b.1(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A.1(a.0(c.0(c.0(x1)))) → A.1(a.0(b.0(b.0(x1))))
A.1(a.0(b.0(b.0(x1)))) → A.1(a.0(x1))
A.1(a.0(b.0(b.1(x1)))) → A.0(a.1(x1))
A.1(a.0(c.0(c.1(x1)))) → A.1(a.0(b.0(b.1(x1))))
The TRS R consists of the following rules:
a.1(a.0(c.0(c.0(x1)))) → c.0(c.0(a.1(a.0(b.0(b.0(x1))))))
b.0(b.0(c.0(c.1(x1)))) → c.0(c.0(b.0(b.0(b.0(b.1(x1))))))
b.0(b.0(c.0(c.0(x1)))) → c.0(c.0(b.0(b.0(b.0(b.0(x1))))))
a.1(a.0(b.0(b.1(x1)))) → b.0(b.0(c.0(c.1(a.0(a.1(x1))))))
a.1(a.0(b.0(b.0(x1)))) → b.0(b.0(c.0(c.0(a.1(a.0(x1))))))
a.1(a.0(c.0(c.1(x1)))) → c.0(c.0(a.1(a.0(b.0(b.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 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
Q DP problem:
The TRS P consists of the following rules:
A.1(a.0(c.0(c.0(x1)))) → A.1(a.0(b.0(b.0(x1))))
A.1(a.0(b.0(b.0(x1)))) → A.1(a.0(x1))
The TRS R consists of the following rules:
a.1(a.0(c.0(c.0(x1)))) → c.0(c.0(a.1(a.0(b.0(b.0(x1))))))
b.0(b.0(c.0(c.1(x1)))) → c.0(c.0(b.0(b.0(b.0(b.1(x1))))))
b.0(b.0(c.0(c.0(x1)))) → c.0(c.0(b.0(b.0(b.0(b.0(x1))))))
a.1(a.0(b.0(b.1(x1)))) → b.0(b.0(c.0(c.1(a.0(a.1(x1))))))
a.1(a.0(b.0(b.0(x1)))) → b.0(b.0(c.0(c.0(a.1(a.0(x1))))))
a.1(a.0(c.0(c.1(x1)))) → c.0(c.0(a.1(a.0(b.0(b.1(x1))))))
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.
No dependency pairs are removed.
The following rules are removed from R:
b.0(b.0(c.0(c.1(x1)))) → c.0(c.0(b.0(b.0(b.0(b.1(x1))))))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.1(x1)) = x1
POL(a.0(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = x1
POL(c.1(x1)) = 1 + x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
A.1(a.0(c.0(c.0(x1)))) → A.1(a.0(b.0(b.0(x1))))
A.1(a.0(b.0(b.0(x1)))) → A.1(a.0(x1))
The TRS R consists of the following rules:
b.0(b.0(c.0(c.0(x1)))) → c.0(c.0(b.0(b.0(b.0(b.0(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.1(a.0(c.0(c.0(x1)))) → A.1(a.0(b.0(b.0(x1))))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.1(x1)) = x1
POL(a.0(x1)) = x1
POL(b.0(x1)) = x1
POL(c.0(x1)) = 1 + x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ UsableRulesReductionPairsProof
Q DP problem:
The TRS P consists of the following rules:
A.1(a.0(b.0(b.0(x1)))) → A.1(a.0(x1))
The TRS R consists of the following rules:
b.0(b.0(c.0(c.0(x1)))) → c.0(c.0(b.0(b.0(b.0(b.0(x1))))))
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.1(a.0(b.0(b.0(x1)))) → A.1(a.0(x1))
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.1(x1)) = x1
POL(a.0(x1)) = x1
POL(b.0(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ 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.