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:
.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
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:
.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
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:
.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
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.