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:

i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x

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, +2(y, z)) -> +12(x, y)
I1(+2(x, y)) -> I1(x)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
I1(+2(x, y)) -> I1(y)
I1(+2(x, y)) -> +12(i1(x), i1(y))

The TRS R consists of the following rules:

i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x

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, +2(y, z)) -> +12(x, y)
I1(+2(x, y)) -> I1(x)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
I1(+2(x, y)) -> I1(y)
I1(+2(x, y)) -> +12(i1(x), i1(y))

The TRS R consists of the following rules:

i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x

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:

+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)

The TRS R consists of the following rules:

i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x

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(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
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)) = 2·x2   
POL(0) = 2   
POL(i1(x1)) = 2   

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:

i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x

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:

I1(+2(x, y)) -> I1(x)
I1(+2(x, y)) -> I1(y)

The TRS R consists of the following rules:

i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x

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.


I1(+2(x, y)) -> I1(x)
I1(+2(x, y)) -> I1(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(I1(x1)) = 2·x1   

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:

i1(0) -> 0
+2(0, y) -> y
+2(x, 0) -> x
i1(i1(x)) -> x
+2(i1(x), x) -> 0
+2(x, i1(x)) -> 0
i1(+2(x, y)) -> +2(i1(x), i1(y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(+2(x, i1(y)), y) -> x
+2(+2(x, y), i1(y)) -> x

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.