Termination Proof using AProVE

Term Rewriting System R:
[y, x]
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(s(x), y) -> +'(x, y)
+'(p(x), y) -> +'(x, y)
MINUS(s(x)) -> MINUS(x)
MINUS(p(x)) -> MINUS(x)
*'(s(x), y) -> +'(*(x, y), y)
*'(s(x), y) -> *'(x, y)
*'(p(x), y) -> +'(*(x, y), minus(y))
*'(p(x), y) -> *'(x, y)
*'(p(x), y) -> MINUS(y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pairs:

+'(p(x), y) -> +'(x, y)
+'(s(x), y) -> +'(x, y)

Rules:

+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))

The following dependency pair can be strictly oriented:

+'(p(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(s(x₁)) = x₁

POL(+'(x₁, x₂)) = x₁

POL(p(x₁)) = 1 + x₁

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 Pair:

+'(s(x), y) -> +'(x, y)

Rules:

+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))

The following dependency pair can be strictly oriented:

+'(s(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(s(x₁)) = 1 + x₁

POL(+'(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Polo

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pair:

Rules:

+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(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:

MINUS(p(x)) -> MINUS(x)
MINUS(s(x)) -> MINUS(x)

Rules:

+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))

The following dependency pair can be strictly oriented:

MINUS(p(x)) -> MINUS(x)

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(MINUS(x₁)) = x₁

POL(s(x₁)) = x₁

POL(p(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 6

             ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

Dependency Pair:

MINUS(s(x)) -> MINUS(x)

Rules:

+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))

The following dependency pair can be strictly oriented:

MINUS(s(x)) -> MINUS(x)

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(MINUS(x₁)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 6

             ↳Polo

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 3

         ↳Polo

Dependency Pair:

Rules:

+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(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:

*'(p(x), y) -> *'(x, y)
*'(s(x), y) -> *'(x, y)

Rules:

+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))

The following dependency pair can be strictly oriented:

*'(p(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(s(x₁)) = x₁

POL(p(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 8

             ↳Polynomial Ordering

Dependency Pair:

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

Rules:

+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))

The following dependency pair can be strictly oriented:

*'(s(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(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 8

             ↳Polo

...

               →DP Problem 9

                 ↳Dependency Graph

Dependency Pair:

Rules:

+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(p(x), y) -> p(+(x, y))
minus(0) -> 0
minus(s(x)) -> p(minus(x))
minus(p(x)) -> s(minus(x))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
*(p(x), y) -> +(*(x, y), minus(y))

Using the Dependency Graph resulted in no new DP problems.

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

POL(s(x₁))	= x₁
POL(+'(x₁, x₂))	= x₁
POL(p(x₁))	= 1 + x₁

POL(MINUS(x₁))	= x₁
POL(s(x₁))	= x₁
POL(p(x₁))	= 1 + x₁

POL(*'(x₁, x₂))	= x₁
POL(s(x₁))	= x₁
POL(p(x₁))	= 1 + x₁