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:
:(:(:(:(C, x), y), z), u) → :(:(x, z), :(:(:(x, y), z), u))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
:(:(:(:(C, x), y), z), u) → :(:(x, z), :(:(:(x, y), z), u))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
:1(:(:(:(C, x), y), z), u) → :1(x, z)
:1(:(:(:(C, x), y), z), u) → :1(:(x, z), :(:(:(x, y), z), u))
:1(:(:(:(C, x), y), z), u) → :1(x, y)
:1(:(:(:(C, x), y), z), u) → :1(:(:(x, y), z), u)
:1(:(:(:(C, x), y), z), u) → :1(:(x, y), z)
The TRS R consists of the following rules:
:(:(:(:(C, x), y), z), u) → :(:(x, z), :(:(:(x, y), z), u))
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:
:1(:(:(:(C, x), y), z), u) → :1(x, z)
:1(:(:(:(C, x), y), z), u) → :1(:(x, z), :(:(:(x, y), z), u))
:1(:(:(:(C, x), y), z), u) → :1(x, y)
:1(:(:(:(C, x), y), z), u) → :1(:(:(x, y), z), u)
:1(:(:(:(C, x), y), z), u) → :1(:(x, y), z)
The TRS R consists of the following rules:
:(:(:(:(C, x), y), z), u) → :(:(x, z), :(:(:(x, y), z), u))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.