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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳AFS

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 AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

+'(x₁, x₂) -> +'(x₁, x₂)
+(x₁, x₂) -> +(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳AFS

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

         ↳AFS

       →DP Problem 2

         ↳Narrowing Transformation

       →DP Problem 3

         ↳AFS

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

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

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

no new Dependency Pairs are created.
The transformation is resulting in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Argument Filtering and 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 AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

+'(x₁, x₂) -> +'(x₁, x₂)
+(x₁, x₂) -> +(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳AFS

           →DP Problem 5

             ↳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