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:
b(c(a(b(b(c(x1)))))) → a(b(c(b(b(c(a(x1)))))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(c(a(b(b(c(x1)))))) → a(b(c(b(b(c(a(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:
B(c(a(b(b(c(x1)))))) → B(c(a(x1)))
B(c(a(b(b(c(x1)))))) → B(b(c(a(x1))))
B(c(a(b(b(c(x1)))))) → B(c(b(b(c(a(x1))))))
The TRS R consists of the following rules:
b(c(a(b(b(c(x1)))))) → a(b(c(b(b(c(a(x1)))))))
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(c(a(b(b(c(x1)))))) → B(c(a(x1)))
B(c(a(b(b(c(x1)))))) → B(b(c(a(x1))))
B(c(a(b(b(c(x1)))))) → B(c(b(b(c(a(x1))))))
The TRS R consists of the following rules:
b(c(a(b(b(c(x1)))))) → a(b(c(b(b(c(a(x1)))))))
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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
B(c(a(b(b(c(x1)))))) → B(c(a(x1)))
The TRS R consists of the following rules:
b(c(a(b(b(c(x1)))))) → a(b(c(b(b(c(a(x1)))))))
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(c(a(b(b(c(x1)))))) → B(c(a(x1)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:
POL(c(x1)) = 1 + x_1
POL(B(x1)) = (2)x_1
POL(a(x1)) = (2)x_1
POL(b(x1)) = x_1
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(c(a(b(b(c(x1)))))) → a(b(c(b(b(c(a(x1)))))))
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.