Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

+2(0, y) -> y
+2(s1(x), 0) -> s1(x)
+2(s1(x), s1(y)) -> s1(+2(s1(x), +2(y, 0)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

+2(0, y) -> y
+2(s1(x), 0) -> s1(x)
+2(s1(x), s1(y)) -> s1(+2(s1(x), +2(y, 0)))

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(0, y) -> y
+2(s1(x), 0) -> s1(x)
+2(s1(x), s1(y)) -> s1(+2(s1(x), +2(y, 0)))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), 0)
+2(s1(x0), s1(x1))


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

+12(s1(x), s1(y)) -> +12(y, 0)
+12(s1(x), s1(y)) -> +12(s1(x), +2(y, 0))

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), 0) -> s1(x)
+2(s1(x), s1(y)) -> s1(+2(s1(x), +2(y, 0)))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), 0)
+2(s1(x0), s1(x1))

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(s1(x), s1(y)) -> +12(y, 0)
+12(s1(x), s1(y)) -> +12(s1(x), +2(y, 0))

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), 0) -> s1(x)
+2(s1(x), s1(y)) -> s1(+2(s1(x), +2(y, 0)))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), 0)
+2(s1(x0), s1(x1))

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
              ↳ QDPAfsSolverProof

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

+12(s1(x), s1(y)) -> +12(s1(x), +2(y, 0))

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), 0) -> s1(x)
+2(s1(x), s1(y)) -> s1(+2(s1(x), +2(y, 0)))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), 0)
+2(s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

+12(s1(x), s1(y)) -> +12(s1(x), +2(y, 0))
Used argument filtering: +12(x1, x2)  =  x2
s1(x1)  =  s1(x1)
+2(x1, x2)  =  +2(x1, x2)
0  =  0
Used ordering: Quasi Precedence: s_1 > +_2 s_1 > 0 


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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), 0) -> s1(x)
+2(s1(x), s1(y)) -> s1(+2(s1(x), +2(y, 0)))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), 0)
+2(s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.