YES
Termination w.r.t. Q proof of ../tpdb/SRS/Waldmann07b/size-12-alpha-3-num-448.srs
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(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
c(b(x1)) → x1
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
c(b(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:
A(x1) → C(x1)
A(b(b(x1))) → A(x1)
A(b(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
c(b(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:
A(x1) → C(x1)
A(b(b(x1))) → A(x1)
A(b(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
c(b(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 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(x1))) → A(x1)
A(b(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
c(b(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(x1))) → A(a(x1)) at position [0] we obtained the following new rules:
A(b(b(b(b(x0))))) → A(b(b(a(a(x0)))))
A(b(b(x0))) → A(b(c(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(b(b(x0))))) → A(b(b(a(a(x0)))))
A(b(b(x0))) → A(b(c(x0)))
A(b(b(x1))) → A(x1)
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
c(b(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.a: x0
b: 1 + x0
A: 0
c: 1 + x0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
A.1(b.0(b.1(x1))) → A.1(x1)
A.0(b.1(b.0(x1))) → A.0(x1)
A.0(b.1(b.0(b.1(b.0(x0))))) → A.0(b.1(b.0(a.0(a.0(x0)))))
A.1(b.0(b.1(b.0(b.1(x0))))) → A.1(b.0(b.1(a.1(a.1(x0)))))
A.1(b.0(b.1(x0))) → A.1(b.0(c.1(x0)))
A.0(b.1(b.0(x0))) → A.0(b.1(c.0(x0)))
The TRS R consists of the following rules:
a.1(x1) → b.0(c.1(x1))
a.0(b.1(b.0(x1))) → b.1(b.0(a.0(a.0(x1))))
a.0(x1) → b.1(c.0(x1))
c.0(b.1(x1)) → x1
c.1(b.0(x1)) → x1
a.1(b.0(b.1(x1))) → b.0(b.1(a.1(a.1(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A.1(b.0(b.1(x1))) → A.1(x1)
A.0(b.1(b.0(x1))) → A.0(x1)
A.0(b.1(b.0(b.1(b.0(x0))))) → A.0(b.1(b.0(a.0(a.0(x0)))))
A.1(b.0(b.1(b.0(b.1(x0))))) → A.1(b.0(b.1(a.1(a.1(x0)))))
A.1(b.0(b.1(x0))) → A.1(b.0(c.1(x0)))
A.0(b.1(b.0(x0))) → A.0(b.1(c.0(x0)))
The TRS R consists of the following rules:
a.1(x1) → b.0(c.1(x1))
a.0(b.1(b.0(x1))) → b.1(b.0(a.0(a.0(x1))))
a.0(x1) → b.1(c.0(x1))
c.0(b.1(x1)) → x1
c.1(b.0(x1)) → x1
a.1(b.0(b.1(x1))) → b.0(b.1(a.1(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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A.1(b.0(b.1(x1))) → A.1(x1)
A.1(b.0(b.1(b.0(b.1(x0))))) → A.1(b.0(b.1(a.1(a.1(x0)))))
A.1(b.0(b.1(x0))) → A.1(b.0(c.1(x0)))
The TRS R consists of the following rules:
a.1(x1) → b.0(c.1(x1))
a.0(b.1(b.0(x1))) → b.1(b.0(a.0(a.0(x1))))
a.0(x1) → b.1(c.0(x1))
c.0(b.1(x1)) → x1
c.1(b.0(x1)) → x1
a.1(b.0(b.1(x1))) → b.0(b.1(a.1(a.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.
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.1(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
POL(c.1(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A.1(b.0(b.1(x1))) → A.1(x1)
A.1(b.0(b.1(b.0(b.1(x0))))) → A.1(b.0(b.1(a.1(a.1(x0)))))
A.1(b.0(b.1(x0))) → A.1(b.0(c.1(x0)))
The TRS R consists of the following rules:
c.1(b.0(x1)) → x1
a.1(x1) → b.0(c.1(x1))
a.1(b.0(b.1(x1))) → b.0(b.1(a.1(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:
A.1(b.0(b.1(x1))) → A.1(x1)
A.1(b.0(b.1(b.0(b.1(x0))))) → A.1(b.0(b.1(a.1(a.1(x0)))))
A.1(b.0(b.1(x0))) → A.1(b.0(c.1(x0)))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.1(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = 1 + x1
POL(c.1(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
The TRS R consists of the following rules:
c.1(b.0(x1)) → x1
a.1(x1) → b.0(c.1(x1))
a.1(b.0(b.1(x1))) → b.0(b.1(a.1(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
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A.0(b.1(b.0(x1))) → A.0(x1)
A.0(b.1(b.0(b.1(b.0(x0))))) → A.0(b.1(b.0(a.0(a.0(x0)))))
A.0(b.1(b.0(x0))) → A.0(b.1(c.0(x0)))
The TRS R consists of the following rules:
a.1(x1) → b.0(c.1(x1))
a.0(b.1(b.0(x1))) → b.1(b.0(a.0(a.0(x1))))
a.0(x1) → b.1(c.0(x1))
c.0(b.1(x1)) → x1
c.1(b.0(x1)) → x1
a.1(b.0(b.1(x1))) → b.0(b.1(a.1(a.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.
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = x1
POL(a.0(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A.0(b.1(b.0(x1))) → A.0(x1)
A.0(b.1(b.0(b.1(b.0(x0))))) → A.0(b.1(b.0(a.0(a.0(x0)))))
A.0(b.1(b.0(x0))) → A.0(b.1(c.0(x0)))
The TRS R consists of the following rules:
a.0(b.1(b.0(x1))) → b.1(b.0(a.0(a.0(x1))))
a.0(x1) → b.1(c.0(x1))
c.0(b.1(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.0(b.1(b.0(x1))) → A.0(x1)
A.0(b.1(b.0(b.1(b.0(x0))))) → A.0(b.1(b.0(a.0(a.0(x0)))))
A.0(b.1(b.0(x0))) → A.0(b.1(c.0(x0)))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = x1
POL(a.0(x1)) = x1
POL(b.0(x1)) = 1 + x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a.0(b.1(b.0(x1))) → b.1(b.0(a.0(a.0(x1))))
a.0(x1) → b.1(c.0(x1))
c.0(b.1(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 have reversed the following QTRS:
The set of rules R is
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
c(b(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
b(c(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(x) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
b(c(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
c(b(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
b(c(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(x) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
b(c(x)) → x
Q is empty.