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(b(c(x1)))
c(b(b(x1))) → a(b(c(x1)))
c(c(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(b(c(x1)))
c(b(b(x1))) → a(b(c(x1)))
c(c(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:
C(b(b(x1))) → C(x1)
A(x1) → C(x1)
C(b(b(x1))) → A(b(c(x1)))
The TRS R consists of the following rules:
a(x1) → b(b(c(x1)))
c(b(b(x1))) → a(b(c(x1)))
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Instantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(x1))) → C(x1)
A(x1) → C(x1)
C(b(b(x1))) → A(b(c(x1)))
The TRS R consists of the following rules:
a(x1) → b(b(c(x1)))
c(b(b(x1))) → a(b(c(x1)))
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule A(x1) → C(x1) we obtained the following new rules:
A(b(y_1)) → C(b(y_1))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Instantiation
↳ QDP
↳ SemLabProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(x1))) → C(x1)
A(b(y_1)) → C(b(y_1))
C(b(b(x1))) → A(b(c(x1)))
The TRS R consists of the following rules:
a(x1) → b(b(c(x1)))
c(b(b(x1))) → a(b(c(x1)))
c(c(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.C: 0
c: 1 + x0
a: 1 + x0
A: 0
b: 1 + x0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
C.0(b.1(b.0(x1))) → A.0(b.1(c.0(x1)))
A.1(b.0(y_1)) → C.1(b.0(y_1))
C.0(b.1(b.0(x1))) → C.0(x1)
C.1(b.0(b.1(x1))) → C.1(x1)
A.0(b.1(y_1)) → C.0(b.1(y_1))
C.1(b.0(b.1(x1))) → A.1(b.0(c.1(x1)))
The TRS R consists of the following rules:
c.0(b.1(b.0(x1))) → a.0(b.1(c.0(x1)))
a.0(x1) → b.0(b.1(c.0(x1)))
c.0(c.1(x1)) → x1
c.1(b.0(b.1(x1))) → a.1(b.0(c.1(x1)))
c.1(c.0(x1)) → x1
a.1(x1) → b.1(b.0(c.1(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Instantiation
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C.0(b.1(b.0(x1))) → A.0(b.1(c.0(x1)))
A.1(b.0(y_1)) → C.1(b.0(y_1))
C.0(b.1(b.0(x1))) → C.0(x1)
C.1(b.0(b.1(x1))) → C.1(x1)
A.0(b.1(y_1)) → C.0(b.1(y_1))
C.1(b.0(b.1(x1))) → A.1(b.0(c.1(x1)))
The TRS R consists of the following rules:
c.0(b.1(b.0(x1))) → a.0(b.1(c.0(x1)))
a.0(x1) → b.0(b.1(c.0(x1)))
c.0(c.1(x1)) → x1
c.1(b.0(b.1(x1))) → a.1(b.0(c.1(x1)))
c.1(c.0(x1)) → x1
a.1(x1) → b.1(b.0(c.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
↳ Instantiation
↳ 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(y_1)) → C.1(b.0(y_1))
C.1(b.0(b.1(x1))) → C.1(x1)
C.1(b.0(b.1(x1))) → A.1(b.0(c.1(x1)))
The TRS R consists of the following rules:
c.0(b.1(b.0(x1))) → a.0(b.1(c.0(x1)))
a.0(x1) → b.0(b.1(c.0(x1)))
c.0(c.1(x1)) → x1
c.1(b.0(b.1(x1))) → a.1(b.0(c.1(x1)))
c.1(c.0(x1)) → x1
a.1(x1) → b.1(b.0(c.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:
c.1(c.0(x1)) → x1
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.1(x1)) = x1
POL(C.1(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = x1
POL(c.1(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Instantiation
↳ 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(y_1)) → C.1(b.0(y_1))
C.1(b.0(b.1(x1))) → C.1(x1)
C.1(b.0(b.1(x1))) → A.1(b.0(c.1(x1)))
The TRS R consists of the following rules:
c.1(b.0(b.1(x1))) → a.1(b.0(c.1(x1)))
a.1(x1) → b.1(b.0(c.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:
C.1(b.0(b.1(x1))) → C.1(x1)
C.1(b.0(b.1(x1))) → A.1(b.0(c.1(x1)))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.1(x1)) = 1 + x1
POL(C.1(x1)) = 1 + x1
POL(a.1(x1)) = 1 + x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = 1 + x1
POL(c.1(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Instantiation
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A.1(b.0(y_1)) → C.1(b.0(y_1))
The TRS R consists of the following rules:
c.1(b.0(b.1(x1))) → a.1(b.0(c.1(x1)))
a.1(x1) → b.1(b.0(c.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 0 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Instantiation
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C.0(b.1(b.0(x1))) → A.0(b.1(c.0(x1)))
C.0(b.1(b.0(x1))) → C.0(x1)
A.0(b.1(y_1)) → C.0(b.1(y_1))
The TRS R consists of the following rules:
c.0(b.1(b.0(x1))) → a.0(b.1(c.0(x1)))
a.0(x1) → b.0(b.1(c.0(x1)))
c.0(c.1(x1)) → x1
c.1(b.0(b.1(x1))) → a.1(b.0(c.1(x1)))
c.1(c.0(x1)) → x1
a.1(x1) → b.1(b.0(c.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:
c.0(c.1(x1)) → x1
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = x1
POL(C.0(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)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Instantiation
↳ 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:
C.0(b.1(b.0(x1))) → A.0(b.1(c.0(x1)))
C.0(b.1(b.0(x1))) → C.0(x1)
A.0(b.1(y_1)) → C.0(b.1(y_1))
The TRS R consists of the following rules:
c.0(b.1(b.0(x1))) → a.0(b.1(c.0(x1)))
a.0(x1) → b.0(b.1(c.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:
C.0(b.1(b.0(x1))) → C.0(x1)
A.0(b.1(y_1)) → C.0(b.1(y_1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = 1 + x1
POL(C.0(x1)) = x1
POL(a.0(x1)) = 1 + x1
POL(b.0(x1)) = 1 + x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Instantiation
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C.0(b.1(b.0(x1))) → A.0(b.1(c.0(x1)))
The TRS R consists of the following rules:
c.0(b.1(b.0(x1))) → a.0(b.1(c.0(x1)))
a.0(x1) → b.0(b.1(c.0(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.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(b(c(x1)))
c(b(b(x1))) → a(b(c(x1)))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(b(x)))
b(b(c(x))) → c(b(a(x)))
c(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(b(x)))
b(b(c(x))) → c(b(a(x)))
c(c(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(b(c(x1)))
c(b(b(x1))) → a(b(c(x1)))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(b(x)))
b(b(c(x))) → c(b(a(x)))
c(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(b(x)))
b(b(c(x))) → c(b(a(x)))
c(c(x)) → x
Q is empty.