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:
:(x, x) → e
:(x, e) → x
i(:(x, y)) → :(y, x)
:(:(x, y), z) → :(x, :(z, i(y)))
:(e, x) → i(x)
i(i(x)) → x
i(e) → e
:(x, :(y, i(x))) → i(y)
:(x, :(y, :(i(x), z))) → :(i(z), y)
:(i(x), :(y, x)) → i(y)
:(i(x), :(y, :(x, z))) → :(i(z), y)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
:(x, x) → e
:(x, e) → x
i(:(x, y)) → :(y, x)
:(:(x, y), z) → :(x, :(z, i(y)))
:(e, x) → i(x)
i(i(x)) → x
i(e) → e
:(x, :(y, i(x))) → i(y)
:(x, :(y, :(i(x), z))) → :(i(z), y)
:(i(x), :(y, x)) → i(y)
:(i(x), :(y, :(x, z))) → :(i(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:
:1(x, :(y, i(x))) → I(y)
:1(e, x) → I(x)
:1(:(x, y), z) → I(y)
:1(:(x, y), z) → :1(z, i(y))
:1(x, :(y, :(i(x), z))) → I(z)
:1(i(x), :(y, x)) → I(y)
:1(i(x), :(y, :(x, z))) → I(z)
:1(:(x, y), z) → :1(x, :(z, i(y)))
I(:(x, y)) → :1(y, x)
:1(x, :(y, :(i(x), z))) → :1(i(z), y)
:1(i(x), :(y, :(x, z))) → :1(i(z), y)
The TRS R consists of the following rules:
:(x, x) → e
:(x, e) → x
i(:(x, y)) → :(y, x)
:(:(x, y), z) → :(x, :(z, i(y)))
:(e, x) → i(x)
i(i(x)) → x
i(e) → e
:(x, :(y, i(x))) → i(y)
:(x, :(y, :(i(x), z))) → :(i(z), y)
:(i(x), :(y, x)) → i(y)
:(i(x), :(y, :(x, z))) → :(i(z), y)
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:
:1(x, :(y, i(x))) → I(y)
:1(e, x) → I(x)
:1(:(x, y), z) → I(y)
:1(:(x, y), z) → :1(z, i(y))
:1(x, :(y, :(i(x), z))) → I(z)
:1(i(x), :(y, x)) → I(y)
:1(i(x), :(y, :(x, z))) → I(z)
:1(:(x, y), z) → :1(x, :(z, i(y)))
I(:(x, y)) → :1(y, x)
:1(x, :(y, :(i(x), z))) → :1(i(z), y)
:1(i(x), :(y, :(x, z))) → :1(i(z), y)
The TRS R consists of the following rules:
:(x, x) → e
:(x, e) → x
i(:(x, y)) → :(y, x)
:(:(x, y), z) → :(x, :(z, i(y)))
:(e, x) → i(x)
i(i(x)) → x
i(e) → e
:(x, :(y, i(x))) → i(y)
:(x, :(y, :(i(x), z))) → :(i(z), y)
:(i(x), :(y, x)) → i(y)
:(i(x), :(y, :(x, z))) → :(i(z), y)
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:
:1(x, :(y, i(x))) → I(y)
:1(:(x, y), z) → I(y)
:1(e, x) → I(x)
:1(:(x, y), z) → :1(z, i(y))
:1(x, :(y, :(i(x), z))) → I(z)
:1(i(x), :(y, x)) → I(y)
:1(:(x, y), z) → :1(x, :(z, i(y)))
:1(i(x), :(y, :(x, z))) → I(z)
I(:(x, y)) → :1(y, x)
:1(x, :(y, :(i(x), z))) → :1(i(z), y)
:1(i(x), :(y, :(x, z))) → :1(i(z), y)
The TRS R consists of the following rules:
:(x, x) → e
:(x, e) → x
i(:(x, y)) → :(y, x)
:(:(x, y), z) → :(x, :(z, i(y)))
:(e, x) → i(x)
i(i(x)) → x
i(e) → e
:(x, :(y, i(x))) → i(y)
:(x, :(y, :(i(x), z))) → :(i(z), y)
:(i(x), :(y, x)) → i(y)
:(i(x), :(y, :(x, z))) → :(i(z), y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.