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:
w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(x))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
B1(w1(x)) -> W1(b1(x))
B1(w1(x)) -> B1(x)
W1(r1(x)) -> W1(x)
B1(r1(x)) -> B1(x)
The TRS R consists of the following rules:
w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(x))
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:
B1(w1(x)) -> W1(b1(x))
B1(w1(x)) -> B1(x)
W1(r1(x)) -> W1(x)
B1(r1(x)) -> B1(x)
The TRS R consists of the following rules:
w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
W1(r1(x)) -> W1(x)
The TRS R consists of the following rules:
w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(x))
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.
W1(r1(x)) -> W1(x)
Used argument filtering: W1(x1) = x1
r1(x1) = r1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(x))
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
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
B1(w1(x)) -> B1(x)
B1(r1(x)) -> B1(x)
The TRS R consists of the following rules:
w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(x))
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.
B1(r1(x)) -> B1(x)
Used argument filtering: B1(x1) = x1
w1(x1) = x1
r1(x1) = r1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
B1(w1(x)) -> B1(x)
The TRS R consists of the following rules:
w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(x))
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.
B1(w1(x)) -> B1(x)
Used argument filtering: B1(x1) = x1
w1(x1) = w1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
w1(r1(x)) -> r1(w1(x))
b1(r1(x)) -> r1(b1(x))
b1(w1(x)) -> w1(b1(x))
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.