Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
g(x, y) -> x
g(x, y) -> y
f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(0, 1, x) -> F(s(x), x, x)
F(x, y, s(z)) -> F(0, 1, z)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Instantiation Transformation

Dependency Pairs:

F(x, y, s(z)) -> F(0, 1, z)
F(0, 1, x) -> F(s(x), x, x)

Rules:

g(x, y) -> x
g(x, y) -> y
f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

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

F(x, y, s(z)) -> F(0, 1, z)

two new Dependency Pairs are created:

F(0, 1, s(z'')) -> F(0, 1, z'')
F(s(s(z')), s(z'), s(z')) -> F(0, 1, z')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Forward Instantiation Transformation

Dependency Pairs:

F(0, 1, s(z'')) -> F(0, 1, z'')
F(s(s(z')), s(z'), s(z')) -> F(0, 1, z')
F(0, 1, x) -> F(s(x), x, x)

Rules:

g(x, y) -> x
g(x, y) -> y
f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

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

F(0, 1, x) -> F(s(x), x, x)

one new Dependency Pair is created:

F(0, 1, s(z'''')) -> F(s(s(z'''')), s(z''''), s(z''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 3

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(s(z')), s(z'), s(z')) -> F(0, 1, z')
F(0, 1, s(z'''')) -> F(s(s(z'''')), s(z''''), s(z''''))
F(0, 1, s(z'')) -> F(0, 1, z'')

Rules:

g(x, y) -> x
g(x, y) -> y
f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

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

F(0, 1, s(z'')) -> F(0, 1, z'')

two new Dependency Pairs are created:

F(0, 1, s(s(z''''))) -> F(0, 1, s(z''''))
F(0, 1, s(s(z''''''))) -> F(0, 1, s(z''''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(0, 1, s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(s(z''''))) -> F(0, 1, s(z''''))
F(0, 1, s(z'''')) -> F(s(s(z'''')), s(z''''), s(z''''))
F(s(s(z')), s(z'), s(z')) -> F(0, 1, z')

Rules:

g(x, y) -> x
g(x, y) -> y
f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

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

F(s(s(z')), s(z'), s(z')) -> F(0, 1, z')

three new Dependency Pairs are created:

F(s(s(s(z''''''))), s(s(z'''''')), s(s(z''''''))) -> F(0, 1, s(z''''''))
F(s(s(s(s(z'''''')))), s(s(s(z''''''))), s(s(s(z'''''')))) -> F(0, 1, s(s(z'''''')))
F(s(s(s(s(z'''''''')))), s(s(s(z''''''''))), s(s(s(z'''''''')))) -> F(0, 1, s(s(z'''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(s(s(s(z'''''''')))), s(s(s(z''''''''))), s(s(s(z'''''''')))) -> F(0, 1, s(s(z'''''''')))
F(s(s(s(s(z'''''')))), s(s(s(z''''''))), s(s(s(z'''''')))) -> F(0, 1, s(s(z'''''')))
F(0, 1, s(s(z''''))) -> F(0, 1, s(z''''))
F(s(s(s(z''''''))), s(s(z'''''')), s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(z'''')) -> F(s(s(z'''')), s(z''''), s(z''''))
F(0, 1, s(s(z''''''))) -> F(0, 1, s(z''''''))

Rules:

g(x, y) -> x
g(x, y) -> y
f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

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

F(0, 1, s(z'''')) -> F(s(s(z'''')), s(z''''), s(z''''))

three new Dependency Pairs are created:

F(0, 1, s(s(z'''''''''))) -> F(s(s(s(z'''''''''))), s(s(z''''''''')), s(s(z''''''''')))
F(0, 1, s(s(s(z''''''''')))) -> F(s(s(s(s(z''''''''')))), s(s(s(z'''''''''))), s(s(s(z'''''''''))))
F(0, 1, s(s(s(z''''''''''')))) -> F(s(s(s(s(z''''''''''')))), s(s(s(z'''''''''''))), s(s(s(z'''''''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 6

                 ↳Argument Filtering and Ordering

Dependency Pairs:

F(0, 1, s(s(s(z''''''''''')))) -> F(s(s(s(s(z''''''''''')))), s(s(s(z'''''''''''))), s(s(s(z'''''''''''))))
F(s(s(s(s(z'''''')))), s(s(s(z''''''))), s(s(s(z'''''')))) -> F(0, 1, s(s(z'''''')))
F(0, 1, s(s(s(z''''''''')))) -> F(s(s(s(s(z''''''''')))), s(s(s(z'''''''''))), s(s(s(z'''''''''))))
F(s(s(s(z''''''))), s(s(z'''''')), s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(s(z'''''''''))) -> F(s(s(s(z'''''''''))), s(s(z''''''''')), s(s(z''''''''')))
F(0, 1, s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(s(z''''))) -> F(0, 1, s(z''''))
F(s(s(s(s(z'''''''')))), s(s(s(z''''''''))), s(s(s(z'''''''')))) -> F(0, 1, s(s(z'''''''')))

Rules:

g(x, y) -> x
g(x, y) -> y
f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(s(s(s(s(z'''''')))), s(s(s(z''''''))), s(s(s(z'''''')))) -> F(0, 1, s(s(z'''''')))
F(s(s(s(z''''''))), s(s(z'''''')), s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(s(z''''''))) -> F(0, 1, s(z''''''))
F(0, 1, s(s(z''''))) -> F(0, 1, s(z''''))
F(s(s(s(s(z'''''''')))), s(s(s(z''''''''))), s(s(s(z'''''''')))) -> F(0, 1, s(s(z'''''''')))

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

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

F(x₁, x₂, x₃) -> x₃
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳FwdInst

...

               →DP Problem 7

                 ↳Dependency Graph

Dependency Pairs:

F(0, 1, s(s(s(z''''''''''')))) -> F(s(s(s(s(z''''''''''')))), s(s(s(z'''''''''''))), s(s(s(z'''''''''''))))
F(0, 1, s(s(s(z''''''''')))) -> F(s(s(s(s(z''''''''')))), s(s(s(z'''''''''))), s(s(s(z'''''''''))))
F(0, 1, s(s(z'''''''''))) -> F(s(s(s(z'''''''''))), s(s(z''''''''')), s(s(z''''''''')))

Rules:

g(x, y) -> x
g(x, y) -> y
f(0, 1, x) -> f(s(x), x, x)
f(x, y, s(z)) -> s(f(0, 1, z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes