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(b(c(a(b(c(x1)))))) → a(b(b(c(b(c(a(x1)))))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(c(a(b(c(x1)))))) → a(b(b(c(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(b(c(a(b(c(x1)))))) → B(b(c(b(c(a(x1))))))
B(b(c(a(b(c(x1)))))) → B(c(b(c(a(x1)))))
B(b(c(a(b(c(x1)))))) → B(c(a(x1)))
The TRS R consists of the following rules:
b(b(c(a(b(c(x1)))))) → a(b(b(c(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(b(c(a(b(c(x1)))))) → B(b(c(b(c(a(x1))))))
B(b(c(a(b(c(x1)))))) → B(c(b(c(a(x1)))))
B(b(c(a(b(c(x1)))))) → B(c(a(x1)))
The TRS R consists of the following rules:
b(b(c(a(b(c(x1)))))) → a(b(b(c(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 0 SCCs with 3 less nodes.