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))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 1 less node.

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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


+12(s1(x), s1(y)) -> +12(s1(x), +2(y, 0))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
+12(x1, x2)  =  +12(x1, x2)
s1(x1)  =  s1(x1)
+2(x1, x2)  =  +2(x1, x2)
0  =  0

Lexicographic Path Order [19].
Precedence:
[+^12, s1] > 0 > +2


The following usable rules [14] were oriented:

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



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