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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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(x, s1(y)) -> +12(x, y)
+12(x, s1(y)) -> S1(+2(x, y))
S1(+2(0, y)) -> S1(y)

The TRS R consists of the following rules:

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

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(x, s1(y)) -> +12(x, y)
+12(x, s1(y)) -> S1(+2(x, y))
S1(+2(0, y)) -> S1(y)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

S1(+2(0, y)) -> S1(y)

The TRS R consists of the following rules:

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

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 strictly oriented and are deleted.


S1(+2(0, y)) -> S1(y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
S1(x1)  =  S1(x1)
+2(x1, x2)  =  +2(x1, x2)
0  =  0

Lexicographic Path Order [19].
Precedence:
+2 > S1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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 strictly oriented and are deleted.


+12(x, s1(y)) -> +12(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
+12(x1, x2)  =  +11(x2)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > +^11

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

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

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