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:
\(x, x) → e
/(x, x) → e
.(e, x) → x
.(x, e) → x
\(e, x) → x
/(x, e) → x
.(x, \(x, y)) → y
.(/(y, x), x) → y
\(x, .(x, y)) → y
/(.(y, x), x) → y
/(x, \(y, x)) → y
\(/(x, y), x) → y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
\(x, x) → e
/(x, x) → e
.(e, x) → x
.(x, e) → x
\(e, x) → x
/(x, e) → x
.(x, \(x, y)) → y
.(/(y, x), x) → y
\(x, .(x, y)) → y
/(.(y, x), x) → y
/(x, \(y, x)) → y
\(/(x, y), x) → y
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
\(x, x) → e
/(x, x) → e
.(e, x) → x
.(x, e) → x
\(e, x) → x
/(x, e) → x
.(x, \(x, y)) → y
.(/(y, x), x) → y
\(x, .(x, y)) → y
/(.(y, x), x) → y
/(x, \(y, x)) → y
\(/(x, y), x) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
\(x, x) → e
/(x, x) → e
.(e, x) → x
.(x, e) → x
\(e, x) → x
/(x, e) → x
.(x, \(x, y)) → y
.(/(y, x), x) → y
\(x, .(x, y)) → y
/(.(y, x), x) → y
/(x, \(y, x)) → y
\(/(x, y), x) → y
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.