Termination w.r.t. Q of the following Term Rewriting System could be disproven:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(x, y, f(z, u, v)) → f(f(x, y, z), u, f(x, y, v))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(x, y, f(z, u, v)) → f(f(x, y, z), u, f(x, y, v))

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:

F(x, y, f(z, u, v)) → F(x, y, z)
F(x, y, f(z, u, v)) → F(f(x, y, z), u, f(x, y, v))
F(x, y, f(z, u, v)) → F(x, y, v)

The TRS R consists of the following rules:

f(x, y, f(z, u, v)) → f(f(x, y, z), u, f(x, y, v))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ NonTerminationProof

Q DP problem:
The TRS P consists of the following rules:

F(x, y, f(z, u, v)) → F(x, y, z)
F(x, y, f(z, u, v)) → F(f(x, y, z), u, f(x, y, v))
F(x, y, f(z, u, v)) → F(x, y, v)

The TRS R consists of the following rules:

f(x, y, f(z, u, v)) → f(f(x, y, z), u, f(x, y, v))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

F(x, y, f(z, u, v)) → F(x, y, z)
F(x, y, f(z, u, v)) → F(f(x, y, z), u, f(x, y, v))
F(x, y, f(z, u, v)) → F(x, y, v)

The TRS R consists of the following rules:

f(x, y, f(z, u, v)) → f(f(x, y, z), u, f(x, y, v))


s = F(x, y, f(z, u, v)) evaluates to t =F(f(x, y, z), u, f(x, y, v))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(x, y, f(z, u, v)) to F(f(x, y, z), u, f(x, y, v)).