MAYBE
Termination w.r.t. Q proof of ../tpdb/TRS/secret05/teparla3.trs
Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
f(x, f(x, y)) → f(f(f(x, a), a), a)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
f(x, f(x, y)) → f(f(f(x, a), a), a)
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:
F(x, f(x, y)) → F(f(x, a), a)
F(x, f(x, y)) → F(f(f(x, a), a), a)
F(y, f(x, f(a, x))) → F(x, a)
F(y, f(x, f(a, x))) → F(a, f(x, a))
F(y, f(x, f(a, x))) → F(a, y)
F(x, f(x, y)) → F(x, a)
F(y, f(x, f(a, x))) → F(f(a, f(x, a)), f(a, y))
The TRS R consists of the following rules:
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
f(x, f(x, y)) → f(f(f(x, a), a), a)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(x, f(x, y)) → F(f(x, a), a)
F(x, f(x, y)) → F(f(f(x, a), a), a)
F(y, f(x, f(a, x))) → F(x, a)
F(y, f(x, f(a, x))) → F(a, f(x, a))
F(y, f(x, f(a, x))) → F(a, y)
F(x, f(x, y)) → F(x, a)
F(y, f(x, f(a, x))) → F(f(a, f(x, a)), f(a, y))
The TRS R consists of the following rules:
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
f(x, f(x, y)) → f(f(f(x, a), a), a)
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 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
Q DP problem:
The TRS P consists of the following rules:
F(y, f(x, f(a, x))) → F(a, y)
F(y, f(x, f(a, x))) → F(f(a, f(x, a)), f(a, y))
The TRS R consists of the following rules:
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
f(x, f(x, y)) → f(f(f(x, a), a), a)
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: 1
f: 0
F: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
F.1-0(y, f.1-0(x, f.1-1(a., x))) → F.1-1(a., y)
F.1-0(y, f.0-0(x, f.1-0(a., x))) → F.1-1(a., y)
F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.1-0(a., y)
F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
F.1-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
F.1-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.1-0(a., y)
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
The TRS R consists of the following rules:
f.0-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
f.0-0(x, f.0-0(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
f.1-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
f.1-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
f.1-0(x, f.1-0(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
f.0-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
f.0-0(x, f.0-1(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
f.1-0(x, f.1-1(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ SemLabProof2
Q DP problem:
The TRS P consists of the following rules:
F.1-0(y, f.1-0(x, f.1-1(a., x))) → F.1-1(a., y)
F.1-0(y, f.0-0(x, f.1-0(a., x))) → F.1-1(a., y)
F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.1-0(a., y)
F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
F.1-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
F.1-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.1-0(a., y)
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
The TRS R consists of the following rules:
f.0-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
f.0-0(x, f.0-0(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
f.1-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
f.1-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
f.1-0(x, f.1-0(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
f.0-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
f.0-0(x, f.0-1(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
f.1-0(x, f.1-1(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
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 6 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ SemLabProof2
Q DP problem:
The TRS P consists of the following rules:
F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
The TRS R consists of the following rules:
f.0-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
f.0-0(x, f.0-0(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
f.1-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
f.1-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
f.1-0(x, f.1-0(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
f.0-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
f.0-0(x, f.0-1(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
f.1-0(x, f.1-1(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
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(F.0-0(x1, x2)) = x1 + x2
POL(a.) = 0
POL(f.0-0(x1, x2)) = x1 + x2
POL(f.0-1(x1, x2)) = x1 + x2
POL(f.1-0(x1, x2)) = x1 + x2
POL(f.1-1(x1, x2)) = x1 + x2
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ DependencyGraphProof
↳ SemLabProof2
Q DP problem:
The TRS P consists of the following rules:
F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
The TRS R consists of the following rules:
f.1-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
f.1-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
f.1-0(x, f.1-0(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
f.1-0(x, f.1-1(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
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
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof2
Q DP problem:
The TRS P consists of the following rules:
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
The TRS R consists of the following rules:
f.1-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
f.1-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
f.1-0(x, f.1-0(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
f.1-0(x, f.1-1(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used.
Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(y, f(x, f(a, x))) → F(f(a, f(x, a)), f(a, y))
The TRS R consists of the following rules:
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
f(x, f(x, y)) → f(f(f(x, a), a), a)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.