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:
:2(x, x) -> e
:2(x, e) -> x
i1(:2(x, y)) -> :2(y, x)
:2(:2(x, y), z) -> :2(x, :2(z, i1(y)))
:2(e, x) -> i1(x)
i1(i1(x)) -> x
i1(e) -> e
:2(x, :2(y, i1(x))) -> i1(y)
:2(x, :2(y, :2(i1(x), z))) -> :2(i1(z), y)
:2(i1(x), :2(y, x)) -> i1(y)
:2(i1(x), :2(y, :2(x, z))) -> :2(i1(z), y)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
:2(x, x) -> e
:2(x, e) -> x
i1(:2(x, y)) -> :2(y, x)
:2(:2(x, y), z) -> :2(x, :2(z, i1(y)))
:2(e, x) -> i1(x)
i1(i1(x)) -> x
i1(e) -> e
:2(x, :2(y, i1(x))) -> i1(y)
:2(x, :2(y, :2(i1(x), z))) -> :2(i1(z), y)
:2(i1(x), :2(y, x)) -> i1(y)
:2(i1(x), :2(y, :2(x, z))) -> :2(i1(z), 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:
:12(x, :2(y, i1(x))) -> I1(y)
:12(e, x) -> I1(x)
:12(:2(x, y), z) -> I1(y)
:12(:2(x, y), z) -> :12(z, i1(y))
:12(x, :2(y, :2(i1(x), z))) -> I1(z)
:12(i1(x), :2(y, x)) -> I1(y)
:12(i1(x), :2(y, :2(x, z))) -> I1(z)
:12(:2(x, y), z) -> :12(x, :2(z, i1(y)))
I1(:2(x, y)) -> :12(y, x)
:12(x, :2(y, :2(i1(x), z))) -> :12(i1(z), y)
:12(i1(x), :2(y, :2(x, z))) -> :12(i1(z), y)
The TRS R consists of the following rules:
:2(x, x) -> e
:2(x, e) -> x
i1(:2(x, y)) -> :2(y, x)
:2(:2(x, y), z) -> :2(x, :2(z, i1(y)))
:2(e, x) -> i1(x)
i1(i1(x)) -> x
i1(e) -> e
:2(x, :2(y, i1(x))) -> i1(y)
:2(x, :2(y, :2(i1(x), z))) -> :2(i1(z), y)
:2(i1(x), :2(y, x)) -> i1(y)
:2(i1(x), :2(y, :2(x, z))) -> :2(i1(z), y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
:12(x, :2(y, i1(x))) -> I1(y)
:12(e, x) -> I1(x)
:12(:2(x, y), z) -> I1(y)
:12(:2(x, y), z) -> :12(z, i1(y))
:12(x, :2(y, :2(i1(x), z))) -> I1(z)
:12(i1(x), :2(y, x)) -> I1(y)
:12(i1(x), :2(y, :2(x, z))) -> I1(z)
:12(:2(x, y), z) -> :12(x, :2(z, i1(y)))
I1(:2(x, y)) -> :12(y, x)
:12(x, :2(y, :2(i1(x), z))) -> :12(i1(z), y)
:12(i1(x), :2(y, :2(x, z))) -> :12(i1(z), y)
The TRS R consists of the following rules:
:2(x, x) -> e
:2(x, e) -> x
i1(:2(x, y)) -> :2(y, x)
:2(:2(x, y), z) -> :2(x, :2(z, i1(y)))
:2(e, x) -> i1(x)
i1(i1(x)) -> x
i1(e) -> e
:2(x, :2(y, i1(x))) -> i1(y)
:2(x, :2(y, :2(i1(x), z))) -> :2(i1(z), y)
:2(i1(x), :2(y, x)) -> i1(y)
:2(i1(x), :2(y, :2(x, z))) -> :2(i1(z), y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.