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

(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)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, 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)
-¹(O(x), O(y)) → O¹(-(x, y))
-¹(O(x), O(y)) → -¹(x, y)
-¹(O(x), I(y)) → -¹(-(x, y), I(1))
-¹(O(x), I(y)) → -¹(x, y)
-¹(I(x), O(y)) → -¹(x, y)
-¹(I(x), I(y)) → O¹(-(x, y))
-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)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, 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 3 SCCs with 7 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), I(1))
-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)

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

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

(7) 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)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))

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