Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
A(a(f(x, y))) → A(y)
F(a(x), a(y)) → F(x, y)
A(a(f(x, y))) → A(b(a(y)))
A(a(f(x, y))) → A(b(a(b(a(x)))))
A(a(f(x, y))) → A(x)
F(b(x), b(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → A(b(a(b(a(y)))))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, y))) → A(b(a(x)))
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
A(a(f(x, y))) → A(y)
F(a(x), a(y)) → F(x, y)
A(a(f(x, y))) → A(b(a(y)))
A(a(f(x, y))) → A(b(a(b(a(x)))))
A(a(f(x, y))) → A(x)
F(b(x), b(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → A(b(a(b(a(y)))))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, y))) → A(b(a(x)))
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A(a(f(x, y))) → A(y)
A(a(f(x, y))) → A(b(a(b(a(x)))))
A(a(f(x, y))) → A(b(a(y)))
F(a(x), a(y)) → F(x, y)
A(a(f(x, y))) → A(x)
F(b(x), b(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
A(a(f(x, y))) → A(b(a(b(a(y)))))
A(a(f(x, y))) → A(b(a(x)))
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A(a(f(x, y))) → A(y)
F(a(x), a(y)) → F(x, y)
A(a(f(x, y))) → A(x)
F(b(x), b(y)) → F(x, y)
F(a(x), a(y)) → A(f(x, y))
A(a(f(x, y))) → F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
The TRS R consists of the following rules:
a(a(f(x, y))) → f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
f(a(x), a(y)) → a(f(x, y))
f(b(x), b(y)) → b(f(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.