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)

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)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

       →DP Problem 2

         ↳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)

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

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

six new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

Dependency Pairs:

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

The following dependency pairs can be strictly oriented:

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

         ↳Nar

           →DP Problem 3

             ↳Polo

...

               →DP Problem 4

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

Dependency Pairs:

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

The following dependency pairs can be strictly oriented:

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

         ↳Nar

           →DP Problem 3

             ↳Polo

...

               →DP Problem 5

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳Polo

Dependency Pairs:

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

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Polo

...

               →DP Problem 6

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

Dependency Pair:

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

The following dependency pair can be strictly oriented:

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

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)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Polo

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳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)

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳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)

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

         ↳Nar

       →DP Problem 2

         ↳Polo

           →DP Problem 8

             ↳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)

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

         ↳Nar

       →DP Problem 2

         ↳Polo

           →DP Problem 8

             ↳Polo

...

               →DP Problem 9

                 ↳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)

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