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:
c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(X) -> e
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(X) -> e
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
C1(b1(a1(X))) -> A1(b1(b1(c1(c1(X)))))
C1(b1(a1(X))) -> C1(X)
C1(b1(a1(X))) -> B1(b1(c1(c1(X))))
C1(b1(a1(X))) -> A1(a1(b1(b1(c1(c1(X))))))
C1(b1(a1(X))) -> B1(c1(c1(X)))
C1(b1(a1(X))) -> C1(c1(X))
The TRS R consists of the following rules:
c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(X) -> e
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:
C1(b1(a1(X))) -> A1(b1(b1(c1(c1(X)))))
C1(b1(a1(X))) -> C1(X)
C1(b1(a1(X))) -> B1(b1(c1(c1(X))))
C1(b1(a1(X))) -> A1(a1(b1(b1(c1(c1(X))))))
C1(b1(a1(X))) -> B1(c1(c1(X)))
C1(b1(a1(X))) -> C1(c1(X))
The TRS R consists of the following rules:
c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(X) -> e
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
C1(b1(a1(X))) -> C1(X)
The TRS R consists of the following rules:
c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(X) -> e
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
C1(b1(a1(X))) -> C1(X)
Used argument filtering: C1(x1) = x1
b1(x1) = x1
a1(x1) = a1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
c1(b1(a1(X))) -> a1(a1(b1(b1(c1(c1(X))))))
a1(X) -> e
b1(X) -> e
c1(X) -> e
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.