Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
  ↳6 QDP

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

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.

(6) Obligation:

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.