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:

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 [1,15] we result in the following initial DP problem:
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), y) → O1(*(x, y))
*1(O(x), y) → O1(*(x, y))
*1(I(x), y) → +1(O(*(x, y)), y)
+1(O(x), O(y)) → +1(x, y)
+1(I(x), I(y)) → +1(+(x, y), I(0))
*1(O(x), y) → *1(x, y)
+1(O(x), O(y)) → O1(+(x, y))
+1(I(x), I(y)) → +1(x, y)
*1(I(x), y) → *1(x, y)
+1(I(x), I(y)) → O1(+(+(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
      ↳ DependencyGraphProof

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), y) → O1(*(x, y))
*1(O(x), y) → O1(*(x, y))
*1(I(x), y) → +1(O(*(x, y)), y)
+1(O(x), O(y)) → +1(x, y)
+1(I(x), I(y)) → +1(+(x, y), I(0))
*1(O(x), y) → *1(x, y)
+1(O(x), O(y)) → O1(+(x, y))
+1(I(x), I(y)) → +1(x, y)
*1(I(x), y) → *1(x, y)
+1(I(x), I(y)) → O1(+(+(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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ UsableRulesProof
          ↳ 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(O(x), O(y)) → +1(x, y)
+1(I(x), I(y)) → +1(+(x, y), I(0))
+1(I(x), I(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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ RuleRemovalProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

+(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)))
O(0) → 0

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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

Strictly oriented rules of the TRS R:

+(I(x), I(y)) → O(+(+(x, y), I(0)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(+(x1, x2)) = x1 + x2   
POL(+1(x1, x2)) = 2·x1 + 2·x2   
POL(0) = 0   
POL(I(x1)) = 1 + x1   
POL(O(x1)) = x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ RuleRemovalProof
QDP
                    ↳ UsableRulesProof
          ↳ 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:

+(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))
O(0) → 0

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ RuleRemovalProof
                  ↳ QDP
                    ↳ UsableRulesProof
QDP
                        ↳ QDPSizeChangeProof
          ↳ QDP

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

+1(O(x), O(y)) → +1(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ UsableRulesProof

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

*1(O(x), y) → *1(x, y)
*1(I(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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: