Termination Proof using AProVE

Term Rewriting System R:
[x, y]
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))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(O(x), O(y)) -> O'(+(x, y))
+'(O(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(I(x), O(y)) -> +'(x, y)
+'(I(x), I(y)) -> O'(+(+(x, y), I(0)))
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), I(y)) -> +'(x, y)
*'(O(x), y) -> O'(*(x, y))
*'(O(x), y) -> *'(x, y)
*'(I(x), y) -> +'(O(*(x, y)), y)
*'(I(x), y) -> O'(*(x, y))
*'(I(x), y) -> *'(x, y)
-'(O(x), O(y)) -> O'(-(x, y))
-'(O(x), O(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(-(x, y), I(1))
-'(O(x), I(y)) -> -'(x, y)
-'(I(x), O(y)) -> -'(x, y)
-'(I(x), I(y)) -> O'(-(x, y))
-'(I(x), I(y)) -> -'(x, y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pairs:

+'(I(x), I(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(O(x), O(y)) -> +'(x, y)

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))

The following dependency pairs can be strictly oriented:

+'(I(x), O(y)) -> +'(x, y)
+'(O(x), O(y)) -> +'(x, y)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(I(x₁)) = x₁

POL(0) = 0

POL(O(x₁)) = 1 + x₁

POL(+(x₁, x₂)) = 0

POL(+'(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pairs:

+'(I(x), I(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(O(x), I(y)) -> +'(x, y)

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))

The following dependency pairs can be strictly oriented:

+'(I(x), I(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(I(x₁)) = 1 + x₁

POL(0) = 0

POL(O(x₁)) = 0

POL(+(x₁, x₂)) = 0

POL(+'(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Polo

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pair:

+'(I(x), I(y)) -> +'(+(x, y), I(0))

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))

The following dependency pair can be strictly oriented:

+'(I(x), I(y)) -> +'(+(x, y), I(0))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

+(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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(I(x₁)) = 1 + x₁

POL(0) = 0

POL(O(x₁)) = x₁

POL(+(x₁, x₂)) = x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Polo

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pair:

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))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

Dependency Pairs:

-'(I(x), I(y)) -> -'(x, y)
-'(I(x), O(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(-(x, y), I(1))
-'(O(x), O(y)) -> -'(x, y)

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))

The following dependency pairs can be strictly oriented:

-'(I(x), O(y)) -> -'(x, y)
-'(O(x), O(y)) -> -'(x, y)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(I(x₁)) = x₁

POL(0) = 0

POL(-'(x₁, x₂)) = x₂

POL(1) = 0

POL(O(x₁)) = 1 + x₁

POL(-(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

Dependency Pairs:

-'(I(x), I(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(-(x, y), I(1))

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))

The following dependency pairs can be strictly oriented:

-'(I(x), I(y)) -> -'(x, y)
-'(O(x), I(y)) -> -'(x, y)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(I(x₁)) = 1 + x₁

POL(0) = 0

POL(-'(x₁, x₂)) = x₂

POL(1) = 0

POL(O(x₁)) = 0

POL(-(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Polo

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

Dependency Pair:

-'(O(x), I(y)) -> -'(-(x, y), I(1))

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))

The following dependency pair can be strictly oriented:

-'(O(x), I(y)) -> -'(-(x, y), I(1))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

-(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))
O(0) -> 0

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(I(x₁)) = 1 + x₁

POL(0) = 0

POL(-'(x₁, x₂)) = 1 + x₁

POL(1) = 0

POL(O(x₁)) = 1 + x₁

POL(-(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Polo

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Polo

Dependency Pair:

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))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

Dependency Pairs:

*'(I(x), y) -> *'(x, y)
*'(O(x), y) -> *'(x, y)

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))

The following dependency pair can be strictly oriented:

*'(I(x), y) -> *'(x, y)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(I(x₁)) = 1 + x₁

POL(*'(x₁, x₂)) = x₁

POL(O(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 10

             ↳Polynomial Ordering

Dependency Pair:

*'(O(x), y) -> *'(x, y)

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))

The following dependency pair can be strictly oriented:

*'(O(x), y) -> *'(x, y)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = x₁

POL(O(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 10

             ↳Polo

...

               →DP Problem 11

                 ↳Dependency Graph

Dependency Pair:

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))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes

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

POL(I(x₁))	= 1 + x₁
POL(*'(x₁, x₂))	= x₁
POL(O(x₁))	= x₁