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:

*(X, +(Y, 1)) → +(*(X, +(Y, *(1, 0))), X)
*(X, 1) → X
*(X, 0) → X
*(X, 0) → 0

The set Q consists of the following terms:

*(x0, 1)
*(x0, +(x1, 1))
*(x0, 0)



↳ QTRS
  ↳ DependencyPairsProof

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

*(X, +(Y, 1)) → +(*(X, +(Y, *(1, 0))), X)
*(X, 1) → X
*(X, 0) → X
*(X, 0) → 0

The set Q consists of the following terms:

*(x0, 1)
*(x0, +(x1, 1))
*(x0, 0)


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(X, +(Y, 1)) → *1(X, +(Y, *(1, 0)))
*1(X, +(Y, 1)) → *1(1, 0)

The TRS R consists of the following rules:

*(X, +(Y, 1)) → +(*(X, +(Y, *(1, 0))), X)
*(X, 1) → X
*(X, 0) → X
*(X, 0) → 0

The set Q consists of the following terms:

*(x0, 1)
*(x0, +(x1, 1))
*(x0, 0)

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:

*1(X, +(Y, 1)) → *1(X, +(Y, *(1, 0)))
*1(X, +(Y, 1)) → *1(1, 0)

The TRS R consists of the following rules:

*(X, +(Y, 1)) → +(*(X, +(Y, *(1, 0))), X)
*(X, 1) → X
*(X, 0) → X
*(X, 0) → 0

The set Q consists of the following terms:

*(x0, 1)
*(x0, +(x1, 1))
*(x0, 0)

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:

*1(X, +(Y, 1)) → *1(X, +(Y, *(1, 0)))

The TRS R consists of the following rules:

*(X, +(Y, 1)) → +(*(X, +(Y, *(1, 0))), X)
*(X, 1) → X
*(X, 0) → X
*(X, 0) → 0

The set Q consists of the following terms:

*(x0, 1)
*(x0, +(x1, 1))
*(x0, 0)

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
              ↳ Narrowing
          ↳ UsableRulesProof

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

*1(X, +(Y, 1)) → *1(X, +(Y, *(1, 0)))

The TRS R consists of the following rules:

*(X, 0) → X
*(X, 0) → 0

The set Q consists of the following terms:

*(x0, 1)
*(x0, +(x1, 1))
*(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule *1(X, +(Y, 1)) → *1(X, +(Y, *(1, 0))) at position [1,1] we obtained the following new rules:

*1(y0, +(y1, 1)) → *1(y0, +(y1, 0))
*1(y0, +(y1, 1)) → *1(y0, +(y1, 1))



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

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

*1(y0, +(y1, 1)) → *1(y0, +(y1, 0))
*1(y0, +(y1, 1)) → *1(y0, +(y1, 1))

The TRS R consists of the following rules:

*(X, 0) → X
*(X, 0) → 0

The set Q consists of the following terms:

*(x0, 1)
*(x0, +(x1, 1))
*(x0, 0)

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
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
          ↳ UsableRulesProof

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

*1(y0, +(y1, 1)) → *1(y0, +(y1, 1))

The TRS R consists of the following rules:

*(X, 0) → X
*(X, 0) → 0

The set Q consists of the following terms:

*(x0, 1)
*(x0, +(x1, 1))
*(x0, 0)

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
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
          ↳ UsableRulesProof

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

*1(y0, +(y1, 1)) → *1(y0, +(y1, 1))

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

*(x0, 1)
*(x0, +(x1, 1))
*(x0, 0)

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.

*(x0, 1)
*(x0, +(x1, 1))
*(x0, 0)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ NonTerminationProof
          ↳ UsableRulesProof

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

*1(y0, +(y1, 1)) → *1(y0, +(y1, 1))

R is empty.
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:

*1(y0, +(y1, 1)) → *1(y0, +(y1, 1))

The TRS R consists of the following rules:none


s = *1(y0, +(y1, 1)) evaluates to t =*1(y0, +(y1, 1))

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 *^1(y0, +(y1, 1)) to *^1(y0, +(y1, 1)).




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
              ↳ NonTerminationProof

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

*1(X, +(Y, 1)) → *1(X, +(Y, *(1, 0)))

The TRS R consists of the following rules:

*(X, 0) → X
*(X, 0) → 0

The set Q consists of the following terms:

*(x0, 1)
*(x0, +(x1, 1))
*(x0, 0)

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 narrowing to the left:

The TRS P consists of the following rules:

*1(X, +(Y, 1)) → *1(X, +(Y, *(1, 0)))

The TRS R consists of the following rules:

*(X, 0) → X
*(X, 0) → 0


s = *1(X, +(Y, *(1, 0))) evaluates to t =*1(X, +(Y, *(1, 0)))

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



Rewriting sequence

*1(X, +(Y, *(1, 0))) → *1(X, +(Y, 1))
with rule *(X', 0) → X' at position [1,1] and matcher [X' / 1]

*1(X, +(Y, 1)) → *1(X, +(Y, *(1, 0)))
with rule *1(X, +(Y, 1)) → *1(X, +(Y, *(1, 0)))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.