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:
.2(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(y), z)) -> z
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
.2(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(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:
.2(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(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:
.2(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(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.