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
  ↳ 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)))

Q is empty.

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)))

Q is empty.
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:

+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)))

Q is empty.
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
  ↳ 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


+12(s1(x), s1(y)) -> +12(s1(x), +2(y, 0))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 1 + x1 + 2·x2   
POL(+12(x1, x2)) = 3·x2   
POL(0) = 0   
POL(s1(x1)) = 3 + 2·x1   

The following usable rules [14] were oriented:

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



↳ 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)))

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