Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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
          ↳ 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.

↳ 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 strictly oriented and are deleted.


*12(I1(x), y) -> *12(x, y)
*12(O1(x), 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(x1)
I1(x1)  =  I1(x1)
O1(x1)  =  O1(x1)

Lexicographic Path Order [19].
Precedence:
trivial

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.