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(b)
b → a

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

f(X) → f(b)
b → a

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

b → a

The TRS R 2 is

f(X) → f(b)

The signature Sigma is {f}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

f(X) → f(b)
b → a

The set Q consists of the following terms:

f(x0)
b


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(b)
F(X) → B

The TRS R consists of the following rules:

f(X) → f(b)
b → a

The set Q consists of the following terms:

f(x0)
b

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(X) → F(b)
F(X) → B

The TRS R consists of the following rules:

f(X) → f(b)
b → a

The set Q consists of the following terms:

f(x0)
b

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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ 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:

f(X) → f(b)
b → a

The set Q consists of the following terms:

f(x0)
b

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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
              ↳ UsableRulesProof

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

F(X) → F(b)

The TRS R consists of the following rules:

b → a

The set Q consists of the following terms:

f(x0)
b

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

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

F(X) → F(b)

The TRS R consists of the following rules:

b → a

The set Q consists of the following terms:

b

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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ MNOCProof
                        ↳ QDP
                          ↳ NonTerminationProof
              ↳ UsableRulesProof

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

F(X) → F(b)

The TRS R consists of the following rules:

b → a

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(b)

The TRS R consists of the following rules:

b → a


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

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(b).




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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof

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

F(X) → F(b)

The TRS R consists of the following rules:

b → a

The set Q consists of the following terms:

f(x0)
b

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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP

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

F(X) → F(b)

The TRS R consists of the following rules:

b → a

The set Q consists of the following terms:

b

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