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:
w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(x))
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:
B(r(x)) → B(x)
W(r(x)) → W(x)
B(w(x)) → W(b(x))
B(w(x)) → B(x)
The TRS R consists of the following rules:
w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(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:
B(r(x)) → B(x)
W(r(x)) → W(x)
B(w(x)) → W(b(x))
B(w(x)) → B(x)
The TRS R consists of the following rules:
w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(x))
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 with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
W(r(x)) → W(x)
The TRS R consists of the following rules:
w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
W(r(x)) → W(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:
POL(r(x1)) = 1/4 + (7/2)x_1
POL(W(x1)) = (2)x_1
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(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
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
B(r(x)) → B(x)
B(w(x)) → B(x)
The TRS R consists of the following rules:
w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
B(r(x)) → B(x)
B(w(x)) → B(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:
POL(w(x1)) = 9/4 + x_1
POL(B(x1)) = (1/2)x_1
POL(r(x1)) = 1/2 + (3/2)x_1
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(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.