Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(x, g(y)) -> F(h(x), i(x, y))
F(x, g(y)) -> I(x, y)
I(x, j(y, z)) -> J(g(y), i(x, z))
I(x, j(y, z)) -> I(x, z)
I(h(x), j(j(y, z), 0)) -> J(i(h(x), j(y, z)), i(x, j(y, z)))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
J(g(x), g(y)) -> J(x, y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Inst

Dependency Pair:

J(g(x), g(y)) -> J(x, y)

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

The following dependency pair can be strictly oriented:

J(g(x), g(y)) -> J(x, y)

There are no usable rules 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:

J(x₁, x₂) -> J(x₁, x₂)
g(x₁) -> g(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Inst

Dependency Pair:

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Inst

Dependency Pairs:

I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(x, j(y, z)) -> I(x, z)

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

The following dependency pairs can be strictly oriented:

I(h(x), j(j(y, z), 0)) -> I(x, j(y, z))
I(h(x), j(j(y, z), 0)) -> I(h(x), j(y, z))
I(x, j(y, z)) -> I(x, z)

The following usable rule w.r.t. to the AFS can be oriented:

j(g(x), g(y)) -> g(j(x, y))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

j > g

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

I(x₁, x₂) -> I(x₁, x₂)
h(x₁) -> h(x₁)
j(x₁, x₂) -> j(x₁, x₂)
g(x₁) -> g(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳Inst

Dependency Pair:

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Instantiation Transformation

Dependency Pair:

F(x, g(y)) -> F(h(x), i(x, y))

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

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

F(x, g(y)) -> F(h(x), i(x, y))

one new Dependency Pair is created:

F(h(x''''), g(y')) -> F(h(h(x'''')), i(h(x''''), y'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Inst

           →DP Problem 6

             ↳Instantiation Transformation

Dependency Pair:

F(h(x''''), g(y')) -> F(h(h(x'''')), i(h(x''''), y'))

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

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

F(h(x''''), g(y')) -> F(h(h(x'''')), i(h(x''''), y'))

one new Dependency Pair is created:

F(h(h(x'''''')), g(y'')) -> F(h(h(h(x''''''))), i(h(h(x'''''')), y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Inst

           →DP Problem 6

             ↳Inst

...

               →DP Problem 7

                 ↳Instantiation Transformation

Dependency Pair:

F(h(h(x'''''')), g(y'')) -> F(h(h(h(x''''''))), i(h(h(x'''''')), y''))

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

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

F(h(h(x'''''')), g(y'')) -> F(h(h(h(x''''''))), i(h(h(x'''''')), y''))

one new Dependency Pair is created:

F(h(h(h(x''''''''))), g(y''')) -> F(h(h(h(h(x'''''''')))), i(h(h(h(x''''''''))), y'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Inst

           →DP Problem 6

             ↳Inst

...

               →DP Problem 8

                 ↳Instantiation Transformation

Dependency Pair:

F(h(h(h(x''''''''))), g(y''')) -> F(h(h(h(h(x'''''''')))), i(h(h(h(x''''''''))), y'''))

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

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

F(h(h(h(x''''''''))), g(y''')) -> F(h(h(h(h(x'''''''')))), i(h(h(h(x''''''''))), y'''))

one new Dependency Pair is created:

F(h(h(h(h(x'''''''''')))), g(y'''')) -> F(h(h(h(h(h(x''''''''''))))), i(h(h(h(h(x'''''''''')))), y''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Inst

           →DP Problem 6

             ↳Inst

...

               →DP Problem 9

                 ↳Instantiation Transformation

Dependency Pair:

F(h(h(h(h(x'''''''''')))), g(y'''')) -> F(h(h(h(h(h(x''''''''''))))), i(h(h(h(h(x'''''''''')))), y''''))

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

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

F(h(h(h(h(x'''''''''')))), g(y'''')) -> F(h(h(h(h(h(x''''''''''))))), i(h(h(h(h(x'''''''''')))), y''''))

one new Dependency Pair is created:

F(h(h(h(h(h(x''''''''''''))))), g(y''''')) -> F(h(h(h(h(h(h(x'''''''''''')))))), i(h(h(h(h(h(x''''''''''''))))), y'''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Inst

           →DP Problem 6

             ↳Inst

...

               →DP Problem 10

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

F(h(h(h(h(h(x''''''''''''))))), g(y''''')) -> F(h(h(h(h(h(h(x'''''''''''')))))), i(h(h(h(h(h(x''''''''''''))))), y'''''))

Rules:

f(x, g(y)) -> f(h(x), i(x, y))
i(x, j(0, 0)) -> g(0)
i(x, j(y, z)) -> j(g(y), i(x, z))
i(h(x), j(j(y, z), 0)) -> j(i(h(x), j(y, z)), i(x, j(y, z)))
j(g(x), g(y)) -> g(j(x, y))

Termination of R could not be shown.
Duration:
0:00 minutes