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:

O1(0) -> 0
+2(0, x) -> x
+2(x, 0) -> x
+2(O1(x), O1(y)) -> O1(+2(x, y))
+2(O1(x), I1(y)) -> I1(+2(x, y))
+2(I1(x), O1(y)) -> I1(+2(x, y))
+2(I1(x), I1(y)) -> O1(+2(+2(x, y), I1(0)))
*2(0, x) -> 0
*2(x, 0) -> 0
*2(O1(x), y) -> O1(*2(x, y))
*2(I1(x), y) -> +2(O1(*2(x, y)), y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

O1(0) -> 0
+2(0, x) -> x
+2(x, 0) -> x
+2(O1(x), O1(y)) -> O1(+2(x, y))
+2(O1(x), I1(y)) -> I1(+2(x, y))
+2(I1(x), O1(y)) -> I1(+2(x, y))
+2(I1(x), I1(y)) -> O1(+2(+2(x, y), I1(0)))
*2(0, x) -> 0
*2(x, 0) -> 0
*2(O1(x), y) -> O1(*2(x, y))
*2(I1(x), y) -> +2(O1(*2(x, y)), 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(I1(x), y) -> *12(x, y)
+12(I1(x), I1(y)) -> O11(+2(+2(x, y), I1(0)))
+12(O1(x), O1(y)) -> O11(+2(x, y))
+12(I1(x), O1(y)) -> +12(x, y)
+12(O1(x), I1(y)) -> +12(x, y)
*12(I1(x), y) -> +12(O1(*2(x, y)), y)
+12(I1(x), I1(y)) -> +12(+2(x, y), I1(0))
*12(O1(x), y) -> *12(x, y)
+12(I1(x), I1(y)) -> +12(x, y)
*12(I1(x), y) -> O11(*2(x, y))
+12(O1(x), O1(y)) -> +12(x, y)
*12(O1(x), y) -> O11(*2(x, y))

The TRS R consists of the following rules:

O1(0) -> 0
+2(0, x) -> x
+2(x, 0) -> x
+2(O1(x), O1(y)) -> O1(+2(x, y))
+2(O1(x), I1(y)) -> I1(+2(x, y))
+2(I1(x), O1(y)) -> I1(+2(x, y))
+2(I1(x), I1(y)) -> O1(+2(+2(x, y), I1(0)))
*2(0, x) -> 0
*2(x, 0) -> 0
*2(O1(x), y) -> O1(*2(x, y))
*2(I1(x), y) -> +2(O1(*2(x, y)), 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(I1(x), y) -> *12(x, y)
+12(I1(x), I1(y)) -> O11(+2(+2(x, y), I1(0)))
+12(O1(x), O1(y)) -> O11(+2(x, y))
+12(I1(x), O1(y)) -> +12(x, y)
+12(O1(x), I1(y)) -> +12(x, y)
*12(I1(x), y) -> +12(O1(*2(x, y)), y)
+12(I1(x), I1(y)) -> +12(+2(x, y), I1(0))
*12(O1(x), y) -> *12(x, y)
+12(I1(x), I1(y)) -> +12(x, y)
*12(I1(x), y) -> O11(*2(x, y))
+12(O1(x), O1(y)) -> +12(x, y)
*12(O1(x), y) -> O11(*2(x, y))

The TRS R consists of the following rules:

O1(0) -> 0
+2(0, x) -> x
+2(x, 0) -> x
+2(O1(x), O1(y)) -> O1(+2(x, y))
+2(O1(x), I1(y)) -> I1(+2(x, y))
+2(I1(x), O1(y)) -> I1(+2(x, y))
+2(I1(x), I1(y)) -> O1(+2(+2(x, y), I1(0)))
*2(0, x) -> 0
*2(x, 0) -> 0
*2(O1(x), y) -> O1(*2(x, y))
*2(I1(x), y) -> +2(O1(*2(x, y)), 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 5 less nodes.

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

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

+12(I1(x), O1(y)) -> +12(x, y)
+12(O1(x), I1(y)) -> +12(x, y)
+12(I1(x), I1(y)) -> +12(+2(x, y), I1(0))
+12(I1(x), I1(y)) -> +12(x, y)
+12(O1(x), O1(y)) -> +12(x, y)

The TRS R consists of the following rules:

O1(0) -> 0
+2(0, x) -> x
+2(x, 0) -> x
+2(O1(x), O1(y)) -> O1(+2(x, y))
+2(O1(x), I1(y)) -> I1(+2(x, y))
+2(I1(x), O1(y)) -> I1(+2(x, y))
+2(I1(x), I1(y)) -> O1(+2(+2(x, y), I1(0)))
*2(0, x) -> 0
*2(x, 0) -> 0
*2(O1(x), y) -> O1(*2(x, y))
*2(I1(x), y) -> +2(O1(*2(x, y)), 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 oriented strictly and are deleted.


+12(I1(x), O1(y)) -> +12(x, y)
+12(O1(x), I1(y)) -> +12(x, y)
+12(I1(x), I1(y)) -> +12(x, y)
+12(O1(x), O1(y)) -> +12(x, y)
The remaining pairs can at least be oriented weakly.

+12(I1(x), I1(y)) -> +12(+2(x, y), I1(0))
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 0   
POL(+12(x1, x2)) = x2   
POL(0) = 0   
POL(I1(x1)) = 1 + x1   
POL(O1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

+12(I1(x), I1(y)) -> +12(+2(x, y), I1(0))

The TRS R consists of the following rules:

O1(0) -> 0
+2(0, x) -> x
+2(x, 0) -> x
+2(O1(x), O1(y)) -> O1(+2(x, y))
+2(O1(x), I1(y)) -> I1(+2(x, y))
+2(I1(x), O1(y)) -> I1(+2(x, y))
+2(I1(x), I1(y)) -> O1(+2(+2(x, y), I1(0)))
*2(0, x) -> 0
*2(x, 0) -> 0
*2(O1(x), y) -> O1(*2(x, y))
*2(I1(x), y) -> +2(O1(*2(x, y)), 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 oriented strictly and are deleted.


+12(I1(x), I1(y)) -> +12(+2(x, y), I1(0))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = x1 + x2   
POL(+12(x1, x2)) = x1 + x2   
POL(0) = 0   
POL(I1(x1)) = 1 + x1   
POL(O1(x1)) = x1   

The following usable rules [14] were oriented:

+2(I1(x), O1(y)) -> I1(+2(x, y))
+2(O1(x), O1(y)) -> O1(+2(x, y))
+2(0, x) -> x
+2(I1(x), I1(y)) -> O1(+2(+2(x, y), I1(0)))
+2(O1(x), I1(y)) -> I1(+2(x, y))
O1(0) -> 0
+2(x, 0) -> x



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

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

O1(0) -> 0
+2(0, x) -> x
+2(x, 0) -> x
+2(O1(x), O1(y)) -> O1(+2(x, y))
+2(O1(x), I1(y)) -> I1(+2(x, y))
+2(I1(x), O1(y)) -> I1(+2(x, y))
+2(I1(x), I1(y)) -> O1(+2(+2(x, y), I1(0)))
*2(0, x) -> 0
*2(x, 0) -> 0
*2(O1(x), y) -> O1(*2(x, y))
*2(I1(x), y) -> +2(O1(*2(x, y)), 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(I1(x), y) -> *12(x, y)
*12(O1(x), y) -> *12(x, y)

The TRS R consists of the following rules:

O1(0) -> 0
+2(0, x) -> x
+2(x, 0) -> x
+2(O1(x), O1(y)) -> O1(+2(x, y))
+2(O1(x), I1(y)) -> I1(+2(x, y))
+2(I1(x), O1(y)) -> I1(+2(x, y))
+2(I1(x), I1(y)) -> O1(+2(+2(x, y), I1(0)))
*2(0, x) -> 0
*2(x, 0) -> 0
*2(O1(x), y) -> O1(*2(x, y))
*2(I1(x), y) -> +2(O1(*2(x, y)), 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 oriented strictly and are deleted.


*12(I1(x), y) -> *12(x, y)
*12(O1(x), y) -> *12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*12(x1, x2)) = x1   
POL(I1(x1)) = 1 + x1   
POL(O1(x1)) = 1 + 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:

O1(0) -> 0
+2(0, x) -> x
+2(x, 0) -> x
+2(O1(x), O1(y)) -> O1(+2(x, y))
+2(O1(x), I1(y)) -> I1(+2(x, y))
+2(I1(x), O1(y)) -> I1(+2(x, y))
+2(I1(x), I1(y)) -> O1(+2(+2(x, y), I1(0)))
*2(0, x) -> 0
*2(x, 0) -> 0
*2(O1(x), y) -> O1(*2(x, y))
*2(I1(x), y) -> +2(O1(*2(x, y)), 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.