Termination Proof using AProVE

Term Rewriting System R:
[x]
+(p1, p1) -> p2
+(p1, +(p2, p2)) -> p5
+(p5, p5) -> p10
+(+(x, y), z) -> +(x, +(y, z))
+(p1, +(p1, x)) -> +(p2, x)
+(p1, +(p2, +(p2, x))) -> +(p5, x)
+(p2, p1) -> +(p1, p2)
+(p2, +(p1, x)) -> +(p1, +(p2, x))
+(p2, +(p2, p2)) -> +(p1, p5)
+(p2, +(p2, +(p2, x))) -> +(p1, +(p5, x))
+(p5, p1) -> +(p1, p5)
+(p5, +(p1, x)) -> +(p1, +(p5, x))
+(p5, p2) -> +(p2, p5)
+(p5, +(p2, x)) -> +(p2, +(p5, x))
+(p5, +(p5, x)) -> +(p10, x)
+(p10, p1) -> +(p1, p10)
+(p10, +(p1, x)) -> +(p1, +(p10, x))
+(p10, p2) -> +(p2, p10)
+(p10, +(p2, x)) -> +(p2, +(p10, x))
+(p10, p5) -> +(p5, p10)
+(p10, +(p5, x)) -> +(p5, +(p10, x))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
+'(p1, +(p1, x)) -> +'(p2, x)
+'(p1, +(p2, +(p2, x))) -> +'(p5, x)
+'(p2, p1) -> +'(p1, p2)
+'(p2, +(p1, x)) -> +'(p1, +(p2, x))
+'(p2, +(p1, x)) -> +'(p2, x)
+'(p2, +(p2, p2)) -> +'(p1, p5)
+'(p2, +(p2, +(p2, x))) -> +'(p1, +(p5, x))
+'(p2, +(p2, +(p2, x))) -> +'(p5, x)
+'(p5, p1) -> +'(p1, p5)
+'(p5, +(p1, x)) -> +'(p1, +(p5, x))
+'(p5, +(p1, x)) -> +'(p5, x)
+'(p5, p2) -> +'(p2, p5)
+'(p5, +(p2, x)) -> +'(p2, +(p5, x))
+'(p5, +(p2, x)) -> +'(p5, x)
+'(p5, +(p5, x)) -> +'(p10, x)
+'(p10, p1) -> +'(p1, p10)
+'(p10, +(p1, x)) -> +'(p1, +(p10, x))
+'(p10, +(p1, x)) -> +'(p10, x)
+'(p10, p2) -> +'(p2, p10)
+'(p10, +(p2, x)) -> +'(p2, +(p10, x))
+'(p10, +(p2, x)) -> +'(p10, x)
+'(p10, p5) -> +'(p5, p10)
+'(p10, +(p5, x)) -> +'(p5, +(p10, x))
+'(p10, +(p5, x)) -> +'(p10, x)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pair:

+'(p10, +(p2, x)) -> +'(p10, x)

Rules:

+(p1, p1) -> p2
+(p1, +(p2, p2)) -> p5
+(p5, p5) -> p10
+(+(x, y), z) -> +(x, +(y, z))
+(p1, +(p1, x)) -> +(p2, x)
+(p1, +(p2, +(p2, x))) -> +(p5, x)
+(p2, p1) -> +(p1, p2)
+(p2, +(p1, x)) -> +(p1, +(p2, x))
+(p2, +(p2, p2)) -> +(p1, p5)
+(p2, +(p2, +(p2, x))) -> +(p1, +(p5, x))
+(p5, p1) -> +(p1, p5)
+(p5, +(p1, x)) -> +(p1, +(p5, x))
+(p5, p2) -> +(p2, p5)
+(p5, +(p2, x)) -> +(p2, +(p5, x))
+(p5, +(p5, x)) -> +(p10, x)
+(p10, p1) -> +(p1, p10)
+(p10, +(p1, x)) -> +(p1, +(p10, x))
+(p10, p2) -> +(p2, p10)
+(p10, +(p2, x)) -> +(p2, +(p10, x))
+(p10, p5) -> +(p5, p10)
+(p10, +(p5, x)) -> +(p5, +(p10, x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

+'(p10, +(p2, x)) -> +'(p10, x)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(p10) = 0

POL(p2) = 0

POL(+(x₁, x₂)) = 1 + x₂

POL(+'(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pair:

Rules:

+(p1, p1) -> p2
+(p1, +(p2, p2)) -> p5
+(p5, p5) -> p10
+(+(x, y), z) -> +(x, +(y, z))
+(p1, +(p1, x)) -> +(p2, x)
+(p1, +(p2, +(p2, x))) -> +(p5, x)
+(p2, p1) -> +(p1, p2)
+(p2, +(p1, x)) -> +(p1, +(p2, x))
+(p2, +(p2, p2)) -> +(p1, p5)
+(p2, +(p2, +(p2, x))) -> +(p1, +(p5, x))
+(p5, p1) -> +(p1, p5)
+(p5, +(p1, x)) -> +(p1, +(p5, x))
+(p5, p2) -> +(p2, p5)
+(p5, +(p2, x)) -> +(p2, +(p5, x))
+(p5, +(p5, x)) -> +(p10, x)
+(p10, p1) -> +(p1, p10)
+(p10, +(p1, x)) -> +(p1, +(p10, x))
+(p10, p2) -> +(p2, p10)
+(p10, +(p2, x)) -> +(p2, +(p10, x))
+(p10, p5) -> +(p5, p10)
+(p10, +(p5, x)) -> +(p5, +(p10, x))

Strategy:

innermost

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 Pair:

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

Rules:

+(p1, p1) -> p2
+(p1, +(p2, p2)) -> p5
+(p5, p5) -> p10
+(+(x, y), z) -> +(x, +(y, z))
+(p1, +(p1, x)) -> +(p2, x)
+(p1, +(p2, +(p2, x))) -> +(p5, x)
+(p2, p1) -> +(p1, p2)
+(p2, +(p1, x)) -> +(p1, +(p2, x))
+(p2, +(p2, p2)) -> +(p1, p5)
+(p2, +(p2, +(p2, x))) -> +(p1, +(p5, x))
+(p5, p1) -> +(p1, p5)
+(p5, +(p1, x)) -> +(p1, +(p5, x))
+(p5, p2) -> +(p2, p5)
+(p5, +(p2, x)) -> +(p2, +(p5, x))
+(p5, +(p5, x)) -> +(p10, x)
+(p10, p1) -> +(p1, p10)
+(p10, +(p1, x)) -> +(p1, +(p10, x))
+(p10, p2) -> +(p2, p10)
+(p10, +(p2, x)) -> +(p2, +(p10, x))
+(p10, p5) -> +(p5, p10)
+(p10, +(p5, x)) -> +(p5, +(p10, x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(z) = 0

POL(p1) = 0

POL(p5) = 0

POL(p2) = 0

POL(p10) = 0

POL(y) = 0

POL(+(x₁, x₂)) = 1 + x₁

POL(+'(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

Dependency Pair:

Rules:

+(p1, p1) -> p2
+(p1, +(p2, p2)) -> p5
+(p5, p5) -> p10
+(+(x, y), z) -> +(x, +(y, z))
+(p1, +(p1, x)) -> +(p2, x)
+(p1, +(p2, +(p2, x))) -> +(p5, x)
+(p2, p1) -> +(p1, p2)
+(p2, +(p1, x)) -> +(p1, +(p2, x))
+(p2, +(p2, p2)) -> +(p1, p5)
+(p2, +(p2, +(p2, x))) -> +(p1, +(p5, x))
+(p5, p1) -> +(p1, p5)
+(p5, +(p1, x)) -> +(p1, +(p5, x))
+(p5, p2) -> +(p2, p5)
+(p5, +(p2, x)) -> +(p2, +(p5, x))
+(p5, +(p5, x)) -> +(p10, x)
+(p10, p1) -> +(p1, p10)
+(p10, +(p1, x)) -> +(p1, +(p10, x))
+(p10, p2) -> +(p2, p10)
+(p10, +(p2, x)) -> +(p2, +(p10, x))
+(p10, p5) -> +(p5, p10)
+(p10, +(p5, x)) -> +(p5, +(p10, x))

Strategy:

innermost

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 Pair:

+'(p5, +(p2, x)) -> +'(p5, x)

Rules:

+(p1, p1) -> p2
+(p1, +(p2, p2)) -> p5
+(p5, p5) -> p10
+(+(x, y), z) -> +(x, +(y, z))
+(p1, +(p1, x)) -> +(p2, x)
+(p1, +(p2, +(p2, x))) -> +(p5, x)
+(p2, p1) -> +(p1, p2)
+(p2, +(p1, x)) -> +(p1, +(p2, x))
+(p2, +(p2, p2)) -> +(p1, p5)
+(p2, +(p2, +(p2, x))) -> +(p1, +(p5, x))
+(p5, p1) -> +(p1, p5)
+(p5, +(p1, x)) -> +(p1, +(p5, x))
+(p5, p2) -> +(p2, p5)
+(p5, +(p2, x)) -> +(p2, +(p5, x))
+(p5, +(p5, x)) -> +(p10, x)
+(p10, p1) -> +(p1, p10)
+(p10, +(p1, x)) -> +(p1, +(p10, x))
+(p10, p2) -> +(p2, p10)
+(p10, +(p2, x)) -> +(p2, +(p10, x))
+(p10, p5) -> +(p5, p10)
+(p10, +(p5, x)) -> +(p5, +(p10, x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

+'(p5, +(p2, x)) -> +'(p5, x)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(p5) = 0

POL(p2) = 0

POL(+(x₁, x₂)) = 1 + x₂

POL(+'(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 6

             ↳Dependency Graph

Dependency Pair:

Rules:

+(p1, p1) -> p2
+(p1, +(p2, p2)) -> p5
+(p5, p5) -> p10
+(+(x, y), z) -> +(x, +(y, z))
+(p1, +(p1, x)) -> +(p2, x)
+(p1, +(p2, +(p2, x))) -> +(p5, x)
+(p2, p1) -> +(p1, p2)
+(p2, +(p1, x)) -> +(p1, +(p2, x))
+(p2, +(p2, p2)) -> +(p1, p5)
+(p2, +(p2, +(p2, x))) -> +(p1, +(p5, x))
+(p5, p1) -> +(p1, p5)
+(p5, +(p1, x)) -> +(p1, +(p5, x))
+(p5, p2) -> +(p2, p5)
+(p5, +(p2, x)) -> +(p2, +(p5, x))
+(p5, +(p5, x)) -> +(p10, x)
+(p10, p1) -> +(p1, p10)
+(p10, +(p1, x)) -> +(p1, +(p10, x))
+(p10, p2) -> +(p2, p10)
+(p10, +(p2, x)) -> +(p2, +(p10, x))
+(p10, p5) -> +(p5, p10)
+(p10, +(p5, x)) -> +(p5, +(p10, x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(p10)	= 0
POL(p2)	= 0
POL(+(x₁, x₂))	= 1 + x₂
POL(+'(x₁, x₂))	= x₂

POL(z)	= 0
POL(p1)	= 0
POL(p5)	= 0
POL(p2)	= 0
POL(p10)	= 0
POL(y)	= 0
POL(+(x₁, x₂))	= 1 + x₁
POL(+'(x₁, x₂))	= x₁