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:
f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)
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:
F(cons(a, k), y) → F(y, k)
F(empty, cons(a, k)) → F(cons(a, k), k)
The TRS R consists of the following rules:
f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)
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:
F(cons(a, k), y) → F(y, k)
F(empty, cons(a, k)) → F(cons(a, k), k)
The TRS R consists of the following rules:
f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)
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
Q DP problem:
The TRS P consists of the following rules:
F(cons(a, k), y) → F(y, k)
F(empty, cons(a, k)) → F(cons(a, k), k)
The TRS R consists of the following rules:
f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.