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:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(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:

*1(I(x), y) → *1(x, y)
+1(I(x), I(y)) → O1(+(+(x, y), I(0)))
+1(O(x), O(y)) → O1(+(x, y))
+1(I(x), O(y)) → +1(x, y)
+1(O(x), I(y)) → +1(x, y)
*1(I(x), y) → +1(O(*(x, y)), y)
+1(I(x), I(y)) → +1(+(x, y), I(0))
*1(O(x), y) → *1(x, y)
+1(I(x), I(y)) → +1(x, y)
*1(I(x), y) → O1(*(x, y))
+1(O(x), O(y)) → +1(x, y)
*1(O(x), y) → O1(*(x, y))

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

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:

*1(I(x), y) → *1(x, y)
+1(I(x), I(y)) → O1(+(+(x, y), I(0)))
+1(O(x), O(y)) → O1(+(x, y))
+1(I(x), O(y)) → +1(x, y)
+1(O(x), I(y)) → +1(x, y)
*1(I(x), y) → +1(O(*(x, y)), y)
+1(I(x), I(y)) → +1(+(x, y), I(0))
*1(O(x), y) → *1(x, y)
+1(I(x), I(y)) → +1(x, y)
*1(I(x), y) → O1(*(x, y))
+1(O(x), O(y)) → +1(x, y)
*1(O(x), y) → O1(*(x, y))

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

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:

+1(O(x), O(y)) → O1(+(x, y))
+1(O(x), I(y)) → +1(x, y)
+1(I(x), O(y)) → +1(x, y)
+1(I(x), I(y)) → +1(x, y)
*1(O(x), y) → *1(x, y)
*1(I(x), y) → *1(x, y)
+1(I(x), I(y)) → O1(+(+(x, y), I(0)))
*1(I(x), y) → +1(O(*(x, y)), y)
+1(I(x), I(y)) → +1(+(x, y), I(0))
+1(O(x), O(y)) → +1(x, y)
*1(I(x), y) → O1(*(x, y))
*1(O(x), y) → O1(*(x, y))

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

+1(I(x), O(y)) → +1(x, y)
+1(O(x), I(y)) → +1(x, y)
+1(I(x), I(y)) → +1(+(x, y), I(0))
+1(I(x), I(y)) → +1(x, y)
+1(O(x), O(y)) → +1(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(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.


+1(O(x), I(y)) → +1(x, y)
+1(I(x), I(y)) → +1(x, y)
The remaining pairs can at least be oriented weakly.

+1(I(x), O(y)) → +1(x, y)
+1(I(x), I(y)) → +1(+(x, y), I(0))
+1(O(x), O(y)) → +1(x, y)
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x2)
I(x1)  =  I(x1)
O(x1)  =  x1
+(x1, x2)  =  +(x1, x2)
0  =  0

Recursive Path Order [2].
Precedence:
I1 > [+^11, +2, 0]


The following usable rules [14] were oriented: none



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

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

+1(I(x), O(y)) → +1(x, y)
+1(I(x), I(y)) → +1(+(x, y), I(0))
+1(O(x), O(y)) → +1(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(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.

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

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

+1(I(x), I(y)) → +1(+(x, y), I(0))

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(x, y)), y)

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

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

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

+1(I(x), O(y)) → +1(x, y)
+1(O(x), O(y)) → +1(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(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.


+1(I(x), O(y)) → +1(x, y)
The remaining pairs can at least be oriented weakly.

+1(O(x), O(y)) → +1(x, y)
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x1, x2)
I(x1)  =  I(x1)
O(x1)  =  x1

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

+1(O(x), O(y)) → +1(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(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.


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

Recursive Path Order [2].
Precedence:
O1 > +^11


The following usable rules [14] were oriented: none



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

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

*1(I(x), y) → *1(x, y)
*1(O(x), y) → *1(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(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.


*1(I(x), y) → *1(x, y)
The remaining pairs can at least be oriented weakly.

*1(O(x), y) → *1(x, y)
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1, x2)
I(x1)  =  I(x1)
O(x1)  =  x1

Recursive Path Order [2].
Precedence:
[*^12, I1]


The following usable rules [14] were oriented: none



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

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

*1(O(x), y) → *1(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(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.


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

Recursive Path Order [2].
Precedence:
O1 > *^11


The following usable rules [14] were oriented: none



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

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
*(0, x) → 0
*(x, 0) → 0
*(O(x), y) → O(*(x, y))
*(I(x), y) → +(O(*(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.