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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
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), y) -> +12(x, y)
+12(s1(x), y) -> +12(x, s1(y))
Used argument filtering: +12(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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), y) -> s1(+2(x, y))
+2(s1(x), y) -> +2(x, s1(y))

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.