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:

f(+(x, 0)) → f(x)
+(x, +(y, z)) → +(+(x, y), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(+(x, 0)) → f(x)
+(x, +(y, z)) → +(+(x, y), z)

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:

F(+(x, 0)) → F(x)
+1(x, +(y, z)) → +1(x, y)
+1(x, +(y, z)) → +1(+(x, y), z)

The TRS R consists of the following rules:

f(+(x, 0)) → f(x)
+(x, +(y, z)) → +(+(x, y), z)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

F(+(x, 0)) → F(x)
+1(x, +(y, z)) → +1(x, y)
+1(x, +(y, z)) → +1(+(x, y), z)

The TRS R consists of the following rules:

f(+(x, 0)) → f(x)
+(x, +(y, z)) → +(+(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(+(x, 0)) → F(x)
+1(x, +(y, z)) → +1(x, y)
+1(x, +(y, z)) → +1(+(x, y), z)

The TRS R consists of the following rules:

f(+(x, 0)) → f(x)
+(x, +(y, z)) → +(+(x, y), z)

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.

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

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

+1(x, +(y, z)) → +1(x, y)
+1(x, +(y, z)) → +1(+(x, y), z)

The TRS R consists of the following rules:

f(+(x, 0)) → f(x)
+(x, +(y, z)) → +(+(x, y), z)

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.


+1(x, +(y, z)) → +1(x, y)
+1(x, +(y, z)) → +1(+(x, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  x2
+(x1, x2)  =  +(x1, x2)

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

f(+(x, 0)) → f(x)
+(x, +(y, z)) → +(+(x, y), z)

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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

F(+(x, 0)) → F(x)

The TRS R consists of the following rules:

f(+(x, 0)) → f(x)
+(x, +(y, z)) → +(+(x, y), z)

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.


F(+(x, 0)) → F(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1)  =  F(x1)
+(x1, x2)  =  +(x1)
0  =  0

Recursive Path Order [2].
Precedence:
+1 > F1
0 > F1


The following usable rules [14] were oriented: none



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

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

f(+(x, 0)) → f(x)
+(x, +(y, z)) → +(+(x, y), z)

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.