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(c)
c → b

The set Q consists of the following terms:

f(x0)
c



↳ QTRS
  ↳ DependencyPairsProof

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

f(X) → f(c)
c → b

The set Q consists of the following terms:

f(x0)
c


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) → C
F(X) → F(c)

The TRS R consists of the following rules:

f(X) → f(c)
c → b

The set Q consists of the following terms:

f(x0)
c

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

F(X) → C
F(X) → F(c)

The TRS R consists of the following rules:

f(X) → f(c)
c → b

The set Q consists of the following terms:

f(x0)
c

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
          ↳ UsableRulesProof

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

F(X) → F(c)

The TRS R consists of the following rules:

f(X) → f(c)
c → b

The set Q consists of the following terms:

f(x0)
c

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
          ↳ UsableRulesProof

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

F(X) → F(c)

The TRS R consists of the following rules:

c → b

The set Q consists of the following terms:

f(x0)
c

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(x0)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Rewriting
          ↳ UsableRulesProof

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

F(X) → F(c)

The TRS R consists of the following rules:

c → b

The set Q consists of the following terms:

c

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(X) → F(c) at position [0] we obtained the following new rules:

F(X) → F(b)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ UsableRulesProof
          ↳ UsableRulesProof

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

F(X) → F(b)

The TRS R consists of the following rules:

c → b

The set Q consists of the following terms:

c

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
          ↳ UsableRulesProof

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

F(X) → F(b)

R is empty.
The set Q consists of the following terms:

c

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

c



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
          ↳ UsableRulesProof

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

F(X) → F(b)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

F(X) → F(c)

The TRS R consists of the following rules:

c → b

The set Q consists of the following terms:

f(x0)
c

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(x0)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ MNOCProof

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

F(X) → F(c)

The TRS R consists of the following rules:

c → b

The set Q consists of the following terms:

c

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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ MNOCProof
                    ↳ QDP
                      ↳ NonTerminationProof

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

F(X) → F(c)

The TRS R consists of the following rules:

c → b

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(c)

The TRS R consists of the following rules:

c → b


s = F(X) evaluates to t =F(c)

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(c).