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) → f(f(x))

The set Q consists of the following terms:

f(x0)



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

f(x) → f(f(x))

The set Q consists of the following terms:

f(x0)


We have reversed the following QTRS:
The set of rules R is

f(x) → f(f(x))

The set Q is {f(x0)}.
We have obtained the following QTRS:

f(x) → f(f(x))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ DependencyPairsProof

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

f(x) → f(f(x))

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) → F(x)
F(x) → F(f(x))

The TRS R consists of the following rules:

f(x) → f(f(x))

The set Q consists of the following terms:

f(x0)

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ MNOCProof

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

F(x) → F(x)
F(x) → F(f(x))

The TRS R consists of the following rules:

f(x) → f(f(x))

The set Q consists of the following terms:

f(x0)

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ MNOCProof
        ↳ QDP
          ↳ NonTerminationProof

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

F(x) → F(x)
F(x) → F(f(x))

The TRS R consists of the following rules:

f(x) → f(f(x))

Q is empty.
We have to consider all (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) → F(x)
F(x) → F(f(x))

The TRS R consists of the following rules:

f(x) → f(f(x))


s = F(x) evaluates to t =F(x)

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) to F(x).