Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

+2(1, x) -> +2(+2(0, 1), x)
+2(0, x) -> x

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

+2(1, x) -> +2(+2(0, 1), x)
+2(0, x) -> x

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

+2(1, x) -> +2(+2(0, 1), x)
+2(0, x) -> x

The set Q consists of the following terms:

+2(1, x0)
+2(0, x0)


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

+12(1, x) -> +12(+2(0, 1), x)
+12(1, x) -> +12(0, 1)

The TRS R consists of the following rules:

+2(1, x) -> +2(+2(0, 1), x)
+2(0, x) -> x

The set Q consists of the following terms:

+2(1, x0)
+2(0, x0)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

+12(1, x) -> +12(+2(0, 1), x)
+12(1, x) -> +12(0, 1)

The TRS R consists of the following rules:

+2(1, x) -> +2(+2(0, 1), x)
+2(0, x) -> x

The set Q consists of the following terms:

+2(1, x0)
+2(0, x0)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP

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

+12(1, x) -> +12(+2(0, 1), x)

The TRS R consists of the following rules:

+2(1, x) -> +2(+2(0, 1), x)
+2(0, x) -> x

The set Q consists of the following terms:

+2(1, x0)
+2(0, x0)

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