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 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE

(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


-1(O(x), O(y)) → -1(x, y)
-1(I(x), O(y)) → -1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
-1(x1, x2)  =  x2
I(x1)  =  x1
1  =  1
O(x1)  =  O(x1)

Recursive path order with status [RPO].
Precedence:
O1 > 1

Status:
trivial

The following usable rules [FROCOS05] were oriented: none

(7) 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(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.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Combined order from the following AFS and order.
-1(x1, x2)  =  x2
I(x1)  =  I(x1)
1  =  1

Recursive path order with status [RPO].
Precedence:
I1 > 1

Status:
trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

-1(O(x), I(y)) → -1(-(x, y), I(1))

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.

(10) 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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


+1(O(x), O(y)) → +1(x, y)
+1(I(x), O(y)) → +1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  x2
I(x1)  =  x1
O(x1)  =  O(x1)
0  =  0

Recursive path order with status [RPO].
Precedence:
O1 > 0

Status:
trivial

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

+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)

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.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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

Recursive path order with status [RPO].
Precedence:
I1 > 0

Status:
trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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)
-(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.

(15) 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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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

Recursive path order with status [RPO].
Precedence:
trivial

Status:
trivial

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

*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)
-(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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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

Recursive path order with status [RPO].
Precedence:
trivial

Status:
trivial

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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)
-(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.

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) TRUE