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)
-2(x, 0) -> x
-2(0, x) -> 0
-2(O1(x), O1(y)) -> O1(-2(x, y))
-2(O1(x), I1(y)) -> I1(-2(-2(x, y), I1(1)))
-2(I1(x), O1(y)) -> I1(-2(x, y))
-2(I1(x), I1(y)) -> O1(-2(x, 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)
-2(x, 0) -> x
-2(0, x) -> 0
-2(O1(x), O1(y)) -> O1(-2(x, y))
-2(O1(x), I1(y)) -> I1(-2(-2(x, y), I1(1)))
-2(I1(x), O1(y)) -> I1(-2(x, y))
-2(I1(x), I1(y)) -> O1(-2(x, y))

Q is empty.

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

-12(I1(x), I1(y)) -> O11(-2(x, y))
+12(O1(x), O1(y)) -> O11(+2(x, y))
+12(O1(x), I1(y)) -> +12(x, y)
+12(I1(x), O1(y)) -> +12(x, y)
-12(O1(x), I1(y)) -> -12(-2(x, y), I1(1))
+12(I1(x), I1(y)) -> +12(x, y)
*12(O1(x), y) -> *12(x, y)
-12(O1(x), O1(y)) -> O11(-2(x, y))
-12(I1(x), I1(y)) -> -12(x, y)
*12(I1(x), y) -> *12(x, y)
-12(I1(x), O1(y)) -> -12(x, y)
-12(O1(x), I1(y)) -> -12(x, y)
+12(I1(x), I1(y)) -> O11(+2(+2(x, y), I1(0)))
-12(O1(x), O1(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), O1(y)) -> +12(x, y)
*12(I1(x), y) -> O11(*2(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)
-2(x, 0) -> x
-2(0, x) -> 0
-2(O1(x), O1(y)) -> O1(-2(x, y))
-2(O1(x), I1(y)) -> I1(-2(-2(x, y), I1(1)))
-2(I1(x), O1(y)) -> I1(-2(x, y))
-2(I1(x), I1(y)) -> O1(-2(x, 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), I1(y)) -> O11(-2(x, y))
+12(O1(x), O1(y)) -> O11(+2(x, y))
+12(O1(x), I1(y)) -> +12(x, y)
+12(I1(x), O1(y)) -> +12(x, y)
-12(O1(x), I1(y)) -> -12(-2(x, y), I1(1))
+12(I1(x), I1(y)) -> +12(x, y)
*12(O1(x), y) -> *12(x, y)
-12(O1(x), O1(y)) -> O11(-2(x, y))
-12(I1(x), I1(y)) -> -12(x, y)
*12(I1(x), y) -> *12(x, y)
-12(I1(x), O1(y)) -> -12(x, y)
-12(O1(x), I1(y)) -> -12(x, y)
+12(I1(x), I1(y)) -> O11(+2(+2(x, y), I1(0)))
-12(O1(x), O1(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), O1(y)) -> +12(x, y)
*12(I1(x), y) -> O11(*2(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)
-2(x, 0) -> x
-2(0, x) -> 0
-2(O1(x), O1(y)) -> O1(-2(x, y))
-2(O1(x), I1(y)) -> I1(-2(-2(x, y), I1(1)))
-2(I1(x), O1(y)) -> I1(-2(x, y))
-2(I1(x), I1(y)) -> O1(-2(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 7 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

-12(I1(x), I1(y)) -> -12(x, y)
-12(I1(x), O1(y)) -> -12(x, y)
-12(O1(x), I1(y)) -> -12(x, y)
-12(O1(x), O1(y)) -> -12(x, y)
-12(O1(x), I1(y)) -> -12(-2(x, y), I1(1))

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)
-2(x, 0) -> x
-2(0, x) -> 0
-2(O1(x), O1(y)) -> O1(-2(x, y))
-2(O1(x), I1(y)) -> I1(-2(-2(x, y), I1(1)))
-2(I1(x), O1(y)) -> I1(-2(x, y))
-2(I1(x), I1(y)) -> O1(-2(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

-12(I1(x), I1(y)) -> -12(x, y)
-12(I1(x), O1(y)) -> -12(x, y)
-12(O1(x), I1(y)) -> -12(x, y)
-12(O1(x), O1(y)) -> -12(x, y)
Used argument filtering: -12(x1, x2)  =  x2
I1(x1)  =  I1(x1)
O1(x1)  =  O1(x1)
1  =  1
-2(x1, x2)  =  x1
0  =  0
Used ordering: Quasi Precedence: [I_1, O_1] > 1 0 > 1 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

-12(O1(x), I1(y)) -> -12(-2(x, y), I1(1))

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)
-2(x, 0) -> x
-2(0, x) -> 0
-2(O1(x), O1(y)) -> O1(-2(x, y))
-2(O1(x), I1(y)) -> I1(-2(-2(x, y), I1(1)))
-2(I1(x), O1(y)) -> I1(-2(x, y))
-2(I1(x), I1(y)) -> O1(-2(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

-12(O1(x), I1(y)) -> -12(-2(x, y), I1(1))
Used argument filtering: -12(x1, x2)  =  x1
O1(x1)  =  O1(x1)
-2(x1, x2)  =  x1
0  =  0
I1(x1)  =  I1(x1)
Used ordering: Quasi Precedence: [O_1, I_1] 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ 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)
-2(x, 0) -> x
-2(0, x) -> 0
-2(O1(x), O1(y)) -> O1(-2(x, y))
-2(O1(x), I1(y)) -> I1(-2(-2(x, y), I1(1)))
-2(I1(x), O1(y)) -> I1(-2(x, y))
-2(I1(x), I1(y)) -> O1(-2(x, 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
          ↳ 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)
-2(x, 0) -> x
-2(0, x) -> 0
-2(O1(x), O1(y)) -> O1(-2(x, y))
-2(O1(x), I1(y)) -> I1(-2(-2(x, y), I1(1)))
-2(I1(x), O1(y)) -> I1(-2(x, y))
-2(I1(x), I1(y)) -> O1(-2(x, y))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof

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)
-2(x, 0) -> x
-2(0, x) -> 0
-2(O1(x), O1(y)) -> O1(-2(x, y))
-2(O1(x), I1(y)) -> I1(-2(-2(x, y), I1(1)))
-2(I1(x), O1(y)) -> I1(-2(x, y))
-2(I1(x), I1(y)) -> O1(-2(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

*12(O1(x), y) -> *12(x, y)
Used argument filtering: *12(x1, x2)  =  x1
I1(x1)  =  x1
O1(x1)  =  O1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof

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

*12(I1(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)
-2(x, 0) -> x
-2(0, x) -> 0
-2(O1(x), O1(y)) -> O1(-2(x, y))
-2(O1(x), I1(y)) -> I1(-2(-2(x, y), I1(1)))
-2(I1(x), O1(y)) -> I1(-2(x, y))
-2(I1(x), I1(y)) -> O1(-2(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

*12(I1(x), y) -> *12(x, y)
Used argument filtering: *12(x1, x2)  =  x1
I1(x1)  =  I1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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)
-2(x, 0) -> x
-2(0, x) -> 0
-2(O1(x), O1(y)) -> O1(-2(x, y))
-2(O1(x), I1(y)) -> I1(-2(-2(x, y), I1(1)))
-2(I1(x), O1(y)) -> I1(-2(x, y))
-2(I1(x), I1(y)) -> O1(-2(x, 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.