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))
b(b(x1)) → x1
b(c(c(x1))) → c(c(a(b(x1))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(c(x1))
b(b(x1)) → x1
b(c(c(x1))) → c(c(a(b(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) → B(c(x1))
B(c(c(x1))) → A(b(x1))
B(c(c(x1))) → B(x1)
The TRS R consists of the following rules:
a(x1) → b(c(x1))
b(b(x1)) → x1
b(c(c(x1))) → c(c(a(b(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ SemLabProof
Q DP problem:
The TRS P consists of the following rules:
A(x1) → B(c(x1))
B(c(c(x1))) → A(b(x1))
B(c(c(x1))) → B(x1)
The TRS R consists of the following rules:
a(x1) → b(c(x1))
b(b(x1)) → x1
b(c(c(x1))) → c(c(a(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: 1 + x0
B: 0
a: 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:
B.1(c.0(c.1(x1))) → B.1(x1)
A.1(x1) → B.0(c.1(x1))
B.0(c.1(c.0(x1))) → A.1(b.0(x1))
B.0(c.1(c.0(x1))) → B.0(x1)
A.0(x1) → B.1(c.0(x1))
B.1(c.0(c.1(x1))) → A.0(b.1(x1))
The TRS R consists of the following rules:
a.1(x1) → b.0(c.1(x1))
a.0(x1) → b.1(c.0(x1))
b.0(c.1(c.0(x1))) → c.0(c.1(a.1(b.0(x1))))
b.1(c.0(c.1(x1))) → c.1(c.0(a.0(b.1(x1))))
b.1(b.0(x1)) → x1
b.0(b.1(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
B.1(c.0(c.1(x1))) → B.1(x1)
A.1(x1) → B.0(c.1(x1))
B.0(c.1(c.0(x1))) → A.1(b.0(x1))
B.0(c.1(c.0(x1))) → B.0(x1)
A.0(x1) → B.1(c.0(x1))
B.1(c.0(c.1(x1))) → A.0(b.1(x1))
The TRS R consists of the following rules:
a.1(x1) → b.0(c.1(x1))
a.0(x1) → b.1(c.0(x1))
b.0(c.1(c.0(x1))) → c.0(c.1(a.1(b.0(x1))))
b.1(c.0(c.1(x1))) → c.1(c.0(a.0(b.1(x1))))
b.1(b.0(x1)) → x1
b.0(b.1(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 2 SCCs.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A.1(x1) → B.0(c.1(x1))
B.0(c.1(c.0(x1))) → A.1(b.0(x1))
B.0(c.1(c.0(x1))) → B.0(x1)
The TRS R consists of the following rules:
a.1(x1) → b.0(c.1(x1))
a.0(x1) → b.1(c.0(x1))
b.0(c.1(c.0(x1))) → c.0(c.1(a.1(b.0(x1))))
b.1(c.0(c.1(x1))) → c.1(c.0(a.0(b.1(x1))))
b.1(b.0(x1)) → x1
b.0(b.1(x1)) → 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(x1) → B.0(c.1(x1))
B.0(c.1(c.0(x1))) → B.0(x1)
The following rules are removed from R:
b.0(b.1(x1)) → x1
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.1(x1)) = 1 + x1
POL(B.0(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = 1 + x1
POL(c.1(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B.0(c.1(c.0(x1))) → A.1(b.0(x1))
The TRS R consists of the following rules:
b.0(c.1(c.0(x1))) → c.0(c.1(a.1(b.0(x1))))
a.1(x1) → 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
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
Q DP problem:
The TRS P consists of the following rules:
B.1(c.0(c.1(x1))) → B.1(x1)
A.0(x1) → B.1(c.0(x1))
B.1(c.0(c.1(x1))) → A.0(b.1(x1))
The TRS R consists of the following rules:
a.1(x1) → b.0(c.1(x1))
a.0(x1) → b.1(c.0(x1))
b.0(c.1(c.0(x1))) → c.0(c.1(a.1(b.0(x1))))
b.1(c.0(c.1(x1))) → c.1(c.0(a.0(b.1(x1))))
b.1(b.0(x1)) → x1
b.0(b.1(x1)) → 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:
B.1(c.0(c.1(x1))) → B.1(x1)
A.0(x1) → B.1(c.0(x1))
The following rules are removed from R:
b.1(b.0(x1)) → x1
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = 1 + x1
POL(B.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
↳ DependencyPairsProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
B.1(c.0(c.1(x1))) → A.0(b.1(x1))
The TRS R consists of the following rules:
b.1(c.0(c.1(x1))) → c.1(c.0(a.0(b.1(x1))))
a.0(x1) → 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.