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:

+(1, x) → +(+(0, 1), x)
+(0, x) → x

The set Q consists of the following terms:

+(0, x0)
+(1, x0)



↳ QTRS
  ↳ DependencyPairsProof

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

+(1, x) → +(+(0, 1), x)
+(0, x) → x

The set Q consists of the following terms:

+(0, x0)
+(1, x0)


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(1, x) → +1(+(0, 1), x)
+1(1, x) → +1(0, 1)

The TRS R consists of the following rules:

+(1, x) → +(+(0, 1), x)
+(0, x) → x

The set Q consists of the following terms:

+(0, x0)
+(1, x0)

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(1, x) → +1(+(0, 1), x)
+1(1, x) → +1(0, 1)

The TRS R consists of the following rules:

+(1, x) → +(+(0, 1), x)
+(0, x) → x

The set Q consists of the following terms:

+(0, x0)
+(1, x0)

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

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

+1(1, x) → +1(+(0, 1), x)

The TRS R consists of the following rules:

+(1, x) → +(+(0, 1), x)
+(0, x) → x

The set Q consists of the following terms:

+(0, x0)
+(1, x0)

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
              ↳ Rewriting

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

+1(1, x) → +1(+(0, 1), x)

The TRS R consists of the following rules:

+(0, x) → x

The set Q consists of the following terms:

+(0, x0)
+(1, x0)

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

+1(1, x) → +1(1, x)



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

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

+1(1, x) → +1(1, x)

The TRS R consists of the following rules:

+(0, x) → x

The set Q consists of the following terms:

+(0, x0)
+(1, x0)

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
              ↳ Rewriting
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ QReductionProof

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

+1(1, x) → +1(1, x)

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

+(0, x0)
+(1, x0)

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.

+(0, x0)
+(1, x0)



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

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

+1(1, x) → +1(1, x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem. 

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

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

1(x) → 1(x)

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(x) → 1(x)

The TRS R consists of the following rules:none


s = 1(x) evaluates to t =1(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 1(x) to 1(x).