Termination Proof using AProVE

Term Rewriting System R:
[y, x, z]
f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(.(nil, y)) -> F(y)
F(.(.(x, y), z)) -> F(.(x, .(y, z)))
G(.(x, nil)) -> G(x)
G(.(x, .(y, z))) -> G(.(.(x, y), z))

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

F(.(.(x, y), z)) -> F(.(x, .(y, z)))
F(.(nil, y)) -> F(y)

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

F(.(nil, y)) -> F(y)

two new Dependency Pairs are created:

F(.(nil, .(nil, y''))) -> F(.(nil, y''))
F(.(nil, .(.(x'', y''), z''))) -> F(.(.(x'', y''), z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

F(.(nil, .(.(x'', y''), z''))) -> F(.(.(x'', y''), z''))
F(.(nil, .(nil, y''))) -> F(.(nil, y''))
F(.(.(x, y), z)) -> F(.(x, .(y, z)))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

F(.(.(x, y), z)) -> F(.(x, .(y, z)))

three new Dependency Pairs are created:

F(.(.(.(x'', y''), y0), z'')) -> F(.(.(x'', y''), .(y0, z'')))
F(.(.(nil, nil), z')) -> F(.(nil, .(nil, z')))
F(.(.(nil, .(x'''', y'''')), z')) -> F(.(nil, .(.(x'''', y''''), z')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

F(.(.(nil, .(x'''', y'''')), z')) -> F(.(nil, .(.(x'''', y''''), z')))
F(.(nil, .(nil, y''))) -> F(.(nil, y''))
F(.(.(nil, nil), z')) -> F(.(nil, .(nil, z')))
F(.(.(.(x'', y''), y0), z'')) -> F(.(.(x'', y''), .(y0, z'')))
F(.(nil, .(.(x'', y''), z''))) -> F(.(.(x'', y''), z''))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

F(.(nil, .(nil, y''))) -> F(.(nil, y''))

two new Dependency Pairs are created:

F(.(nil, .(nil, .(nil, y'''')))) -> F(.(nil, .(nil, y'''')))
F(.(nil, .(nil, .(.(x'''', y''''), z'''')))) -> F(.(nil, .(.(x'''', y''''), z'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

F(.(nil, .(nil, .(.(x'''', y''''), z'''')))) -> F(.(nil, .(.(x'''', y''''), z'''')))
F(.(nil, .(nil, .(nil, y'''')))) -> F(.(nil, .(nil, y'''')))
F(.(.(nil, nil), z')) -> F(.(nil, .(nil, z')))
F(.(.(.(x'', y''), y0), z'')) -> F(.(.(x'', y''), .(y0, z'')))
F(.(nil, .(.(x'', y''), z''))) -> F(.(.(x'', y''), z''))
F(.(.(nil, .(x'''', y'''')), z')) -> F(.(nil, .(.(x'''', y''''), z')))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

F(.(nil, .(.(x'', y''), z''))) -> F(.(.(x'', y''), z''))

three new Dependency Pairs are created:

F(.(nil, .(.(.(x'''', y''''), y''0), z''''))) -> F(.(.(.(x'''', y''''), y''0), z''''))
F(.(nil, .(.(nil, nil), z''''))) -> F(.(.(nil, nil), z''''))
F(.(nil, .(.(nil, .(x'''''', y'''''')), z''''))) -> F(.(.(nil, .(x'''''', y'''''')), z''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

F(.(nil, .(.(nil, .(x'''''', y'''''')), z''''))) -> F(.(.(nil, .(x'''''', y'''''')), z''''))
F(.(nil, .(.(nil, nil), z''''))) -> F(.(.(nil, nil), z''''))
F(.(.(nil, .(x'''', y'''')), z')) -> F(.(nil, .(.(x'''', y''''), z')))
F(.(nil, .(nil, .(nil, y'''')))) -> F(.(nil, .(nil, y'''')))
F(.(.(nil, nil), z')) -> F(.(nil, .(nil, z')))
F(.(.(.(x'', y''), y0), z'')) -> F(.(.(x'', y''), .(y0, z'')))
F(.(nil, .(.(.(x'''', y''''), y''0), z''''))) -> F(.(.(.(x'''', y''''), y''0), z''''))
F(.(nil, .(nil, .(.(x'''', y''''), z'''')))) -> F(.(nil, .(.(x'''', y''''), z'''')))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

F(.(.(.(x'', y''), y0), z'')) -> F(.(.(x'', y''), .(y0, z'')))

three new Dependency Pairs are created:

F(.(.(.(.(x'''', y''''), y''0), y0''), z'''')) -> F(.(.(.(x'''', y''''), y''0), .(y0'', z'''')))
F(.(.(.(nil, nil), y0'), z'''')) -> F(.(.(nil, nil), .(y0', z'''')))
F(.(.(.(nil, .(x'''''', y'''''')), y0'), z'''')) -> F(.(.(nil, .(x'''''', y'''''')), .(y0', z'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

F(.(.(.(nil, .(x'''''', y'''''')), y0'), z'''')) -> F(.(.(nil, .(x'''''', y'''''')), .(y0', z'''')))
F(.(nil, .(.(nil, nil), z''''))) -> F(.(.(nil, nil), z''''))
F(.(nil, .(nil, .(.(x'''', y''''), z'''')))) -> F(.(nil, .(.(x'''', y''''), z'''')))
F(.(nil, .(nil, .(nil, y'''')))) -> F(.(nil, .(nil, y'''')))
F(.(.(nil, nil), z')) -> F(.(nil, .(nil, z')))
F(.(.(.(nil, nil), y0'), z'''')) -> F(.(.(nil, nil), .(y0', z'''')))
F(.(.(.(.(x'''', y''''), y''0), y0''), z'''')) -> F(.(.(.(x'''', y''''), y''0), .(y0'', z'''')))
F(.(nil, .(.(.(x'''', y''''), y''0), z''''))) -> F(.(.(.(x'''', y''''), y''0), z''''))
F(.(.(nil, .(x'''', y'''')), z')) -> F(.(nil, .(.(x'''', y''''), z')))
F(.(nil, .(.(nil, .(x'''''', y'''''')), z''''))) -> F(.(.(nil, .(x'''''', y'''''')), z''''))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

F(.(.(nil, nil), z')) -> F(.(nil, .(nil, z')))

two new Dependency Pairs are created:

F(.(.(nil, nil), .(nil, y''''''))) -> F(.(nil, .(nil, .(nil, y''''''))))
F(.(.(nil, nil), .(.(x'''''', y''''''), z''''''))) -> F(.(nil, .(nil, .(.(x'''''', y''''''), z''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 8

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

F(.(nil, .(.(nil, .(x'''''', y'''''')), z''''))) -> F(.(.(nil, .(x'''''', y'''''')), z''''))
F(.(.(nil, nil), .(.(x'''''', y''''''), z''''''))) -> F(.(nil, .(nil, .(.(x'''''', y''''''), z''''''))))
F(.(nil, .(.(nil, nil), z''''))) -> F(.(.(nil, nil), z''''))
F(.(nil, .(nil, .(.(x'''', y''''), z'''')))) -> F(.(nil, .(.(x'''', y''''), z'''')))
F(.(nil, .(nil, .(nil, y'''')))) -> F(.(nil, .(nil, y'''')))
F(.(.(nil, nil), .(nil, y''''''))) -> F(.(nil, .(nil, .(nil, y''''''))))
F(.(.(.(nil, nil), y0'), z'''')) -> F(.(.(nil, nil), .(y0', z'''')))
F(.(.(.(.(x'''', y''''), y''0), y0''), z'''')) -> F(.(.(.(x'''', y''''), y''0), .(y0'', z'''')))
F(.(nil, .(.(.(x'''', y''''), y''0), z''''))) -> F(.(.(.(x'''', y''''), y''0), z''''))
F(.(.(nil, .(x'''', y'''')), z')) -> F(.(nil, .(.(x'''', y''''), z')))
F(.(.(.(nil, .(x'''''', y'''''')), y0'), z'''')) -> F(.(.(nil, .(x'''''', y'''''')), .(y0', z'''')))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

F(.(.(nil, .(x'''', y'''')), z')) -> F(.(nil, .(.(x'''', y''''), z')))

three new Dependency Pairs are created:

F(.(.(nil, .(.(x'''''', y''''''), y''''0)), z'')) -> F(.(nil, .(.(.(x'''''', y''''''), y''''0), z'')))
F(.(.(nil, .(nil, nil)), z'')) -> F(.(nil, .(.(nil, nil), z'')))
F(.(.(nil, .(nil, .(x'''''''', y''''''''))), z'')) -> F(.(nil, .(.(nil, .(x'''''''', y'''''''')), z'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 9

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

F(.(.(nil, .(nil, .(x'''''''', y''''''''))), z'')) -> F(.(nil, .(.(nil, .(x'''''''', y'''''''')), z'')))
F(.(.(nil, .(nil, nil)), z'')) -> F(.(nil, .(.(nil, nil), z'')))
F(.(.(.(nil, .(x'''''', y'''''')), y0'), z'''')) -> F(.(.(nil, .(x'''''', y'''''')), .(y0', z'''')))
F(.(.(nil, nil), .(.(x'''''', y''''''), z''''''))) -> F(.(nil, .(nil, .(.(x'''''', y''''''), z''''''))))
F(.(nil, .(.(nil, nil), z''''))) -> F(.(.(nil, nil), z''''))
F(.(nil, .(nil, .(.(x'''', y''''), z'''')))) -> F(.(nil, .(.(x'''', y''''), z'''')))
F(.(nil, .(nil, .(nil, y'''')))) -> F(.(nil, .(nil, y'''')))
F(.(.(nil, nil), .(nil, y''''''))) -> F(.(nil, .(nil, .(nil, y''''''))))
F(.(.(.(nil, nil), y0'), z'''')) -> F(.(.(nil, nil), .(y0', z'''')))
F(.(.(.(.(x'''', y''''), y''0), y0''), z'''')) -> F(.(.(.(x'''', y''''), y''0), .(y0'', z'''')))
F(.(nil, .(.(.(x'''', y''''), y''0), z''''))) -> F(.(.(.(x'''', y''''), y''0), z''''))
F(.(.(nil, .(.(x'''''', y''''''), y''''0)), z'')) -> F(.(nil, .(.(.(x'''''', y''''''), y''''0), z'')))
F(.(nil, .(.(nil, .(x'''''', y'''''')), z''''))) -> F(.(.(nil, .(x'''''', y'''''')), z''''))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(.(nil, .(.(nil, nil), z''''))) -> F(.(.(nil, nil), z''''))
F(.(nil, .(nil, .(.(x'''', y''''), z'''')))) -> F(.(nil, .(.(x'''', y''''), z'''')))
F(.(nil, .(nil, .(nil, y'''')))) -> F(.(nil, .(nil, y'''')))
F(.(nil, .(.(.(x'''', y''''), y''0), z''''))) -> F(.(.(.(x'''', y''''), y''0), z''''))
F(.(nil, .(.(nil, .(x'''''', y'''''')), z''''))) -> F(.(.(nil, .(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(nil) = 0

POL(.(x₁, x₂)) = 1 + x₁ + x₂

POL(F(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 11

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

F(.(.(.(.(x'''', y''''), y''0), y0''), z'''')) -> F(.(.(.(x'''', y''''), y''0), .(y0'', z'''')))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(.(.(.(.(x'''', y''''), y''0), y0''), z'''')) -> F(.(.(.(x'''', y''''), y''0), .(y0'', 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(.(x₁, x₂)) = 1 + x₁

POL(F(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 12

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

Dependency Pairs:

G(.(x, .(y, z))) -> G(.(.(x, y), z))
G(.(x, nil)) -> G(x)

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

G(.(x, nil)) -> G(x)

two new Dependency Pairs are created:

G(.(.(x'', nil), nil)) -> G(.(x'', nil))
G(.(.(x'', .(y'', z'')), nil)) -> G(.(x'', .(y'', z'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 13

             ↳Forward Instantiation Transformation

Dependency Pairs:

G(.(.(x'', .(y'', z'')), nil)) -> G(.(x'', .(y'', z'')))
G(.(.(x'', nil), nil)) -> G(.(x'', nil))
G(.(x, .(y, z))) -> G(.(.(x, y), z))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

G(.(x, .(y, z))) -> G(.(.(x, y), z))

three new Dependency Pairs are created:

G(.(x'', .(y0, .(y'', z'')))) -> G(.(.(x'', y0), .(y'', z'')))
G(.(x', .(nil, nil))) -> G(.(.(x', nil), nil))
G(.(x', .(.(y'''', z''''), nil))) -> G(.(.(x', .(y'''', z'''')), nil))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

Dependency Pairs:

G(.(x', .(.(y'''', z''''), nil))) -> G(.(.(x', .(y'''', z'''')), nil))
G(.(.(x'', nil), nil)) -> G(.(x'', nil))
G(.(x', .(nil, nil))) -> G(.(.(x', nil), nil))
G(.(x'', .(y0, .(y'', z'')))) -> G(.(.(x'', y0), .(y'', z'')))
G(.(.(x'', .(y'', z'')), nil)) -> G(.(x'', .(y'', z'')))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

G(.(.(x'', nil), nil)) -> G(.(x'', nil))

two new Dependency Pairs are created:

G(.(.(.(x'''', nil), nil), nil)) -> G(.(.(x'''', nil), nil))
G(.(.(.(x'''', .(y'''', z'''')), nil), nil)) -> G(.(.(x'''', .(y'''', z'''')), nil))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 15

                 ↳Forward Instantiation Transformation

Dependency Pairs:

G(.(.(.(x'''', .(y'''', z'''')), nil), nil)) -> G(.(.(x'''', .(y'''', z'''')), nil))
G(.(.(.(x'''', nil), nil), nil)) -> G(.(.(x'''', nil), nil))
G(.(x', .(nil, nil))) -> G(.(.(x', nil), nil))
G(.(x'', .(y0, .(y'', z'')))) -> G(.(.(x'', y0), .(y'', z'')))
G(.(.(x'', .(y'', z'')), nil)) -> G(.(x'', .(y'', z'')))
G(.(x', .(.(y'''', z''''), nil))) -> G(.(.(x', .(y'''', z'''')), nil))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

G(.(.(x'', .(y'', z'')), nil)) -> G(.(x'', .(y'', z'')))

three new Dependency Pairs are created:

G(.(.(x'''', .(y''0, .(y'''', z''''))), nil)) -> G(.(x'''', .(y''0, .(y'''', z''''))))
G(.(.(x'''', .(nil, nil)), nil)) -> G(.(x'''', .(nil, nil)))
G(.(.(x'''', .(.(y'''''', z''''''), nil)), nil)) -> G(.(x'''', .(.(y'''''', z''''''), nil)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 16

                 ↳Forward Instantiation Transformation

Dependency Pairs:

G(.(.(x'''', .(.(y'''''', z''''''), nil)), nil)) -> G(.(x'''', .(.(y'''''', z''''''), nil)))
G(.(.(x'''', .(nil, nil)), nil)) -> G(.(x'''', .(nil, nil)))
G(.(x', .(.(y'''', z''''), nil))) -> G(.(.(x', .(y'''', z'''')), nil))
G(.(.(.(x'''', nil), nil), nil)) -> G(.(.(x'''', nil), nil))
G(.(x', .(nil, nil))) -> G(.(.(x', nil), nil))
G(.(x'', .(y0, .(y'', z'')))) -> G(.(.(x'', y0), .(y'', z'')))
G(.(.(x'''', .(y''0, .(y'''', z''''))), nil)) -> G(.(x'''', .(y''0, .(y'''', z''''))))
G(.(.(.(x'''', .(y'''', z'''')), nil), nil)) -> G(.(.(x'''', .(y'''', z'''')), nil))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

G(.(x'', .(y0, .(y'', z'')))) -> G(.(.(x'', y0), .(y'', z'')))

three new Dependency Pairs are created:

G(.(x'''', .(y0'', .(y''0, .(y'''', z''''))))) -> G(.(.(x'''', y0''), .(y''0, .(y'''', z''''))))
G(.(x'''', .(y0', .(nil, nil)))) -> G(.(.(x'''', y0'), .(nil, nil)))
G(.(x'''', .(y0', .(.(y'''''', z''''''), nil)))) -> G(.(.(x'''', y0'), .(.(y'''''', z''''''), nil)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 17

                 ↳Forward Instantiation Transformation

Dependency Pairs:

G(.(x'''', .(y0', .(.(y'''''', z''''''), nil)))) -> G(.(.(x'''', y0'), .(.(y'''''', z''''''), nil)))
G(.(.(x'''', .(nil, nil)), nil)) -> G(.(x'''', .(nil, nil)))
G(.(.(.(x'''', .(y'''', z'''')), nil), nil)) -> G(.(.(x'''', .(y'''', z'''')), nil))
G(.(.(.(x'''', nil), nil), nil)) -> G(.(.(x'''', nil), nil))
G(.(x', .(nil, nil))) -> G(.(.(x', nil), nil))
G(.(x'''', .(y0', .(nil, nil)))) -> G(.(.(x'''', y0'), .(nil, nil)))
G(.(x'''', .(y0'', .(y''0, .(y'''', z''''))))) -> G(.(.(x'''', y0''), .(y''0, .(y'''', z''''))))
G(.(.(x'''', .(y''0, .(y'''', z''''))), nil)) -> G(.(x'''', .(y''0, .(y'''', z''''))))
G(.(x', .(.(y'''', z''''), nil))) -> G(.(.(x', .(y'''', z'''')), nil))
G(.(.(x'''', .(.(y'''''', z''''''), nil)), nil)) -> G(.(x'''', .(.(y'''''', z''''''), nil)))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

G(.(x', .(nil, nil))) -> G(.(.(x', nil), nil))

two new Dependency Pairs are created:

G(.(.(x'''''', nil), .(nil, nil))) -> G(.(.(.(x'''''', nil), nil), nil))
G(.(.(x'''''', .(y'''''', z'''''')), .(nil, nil))) -> G(.(.(.(x'''''', .(y'''''', z'''''')), nil), nil))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 18

                 ↳Forward Instantiation Transformation

Dependency Pairs:

G(.(.(x'''', .(.(y'''''', z''''''), nil)), nil)) -> G(.(x'''', .(.(y'''''', z''''''), nil)))
G(.(.(x'''''', .(y'''''', z'''''')), .(nil, nil))) -> G(.(.(.(x'''''', .(y'''''', z'''''')), nil), nil))
G(.(.(x'''', .(nil, nil)), nil)) -> G(.(x'''', .(nil, nil)))
G(.(.(.(x'''', .(y'''', z'''')), nil), nil)) -> G(.(.(x'''', .(y'''', z'''')), nil))
G(.(.(.(x'''', nil), nil), nil)) -> G(.(.(x'''', nil), nil))
G(.(.(x'''''', nil), .(nil, nil))) -> G(.(.(.(x'''''', nil), nil), nil))
G(.(x'''', .(y0', .(nil, nil)))) -> G(.(.(x'''', y0'), .(nil, nil)))
G(.(x'''', .(y0'', .(y''0, .(y'''', z''''))))) -> G(.(.(x'''', y0''), .(y''0, .(y'''', z''''))))
G(.(.(x'''', .(y''0, .(y'''', z''''))), nil)) -> G(.(x'''', .(y''0, .(y'''', z''''))))
G(.(x', .(.(y'''', z''''), nil))) -> G(.(.(x', .(y'''', z'''')), nil))
G(.(x'''', .(y0', .(.(y'''''', z''''''), nil)))) -> G(.(.(x'''', y0'), .(.(y'''''', z''''''), nil)))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

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

G(.(x', .(.(y'''', z''''), nil))) -> G(.(.(x', .(y'''', z'''')), nil))

three new Dependency Pairs are created:

G(.(x'', .(.(y''''0, .(y'''''', z'''''')), nil))) -> G(.(.(x'', .(y''''0, .(y'''''', z''''''))), nil))
G(.(x'', .(.(nil, nil), nil))) -> G(.(.(x'', .(nil, nil)), nil))
G(.(x'', .(.(.(y'''''''', z''''''''), nil), nil))) -> G(.(.(x'', .(.(y'''''''', z''''''''), nil)), nil))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 19

                 ↳Polynomial Ordering

Dependency Pairs:

G(.(x'', .(.(.(y'''''''', z''''''''), nil), nil))) -> G(.(.(x'', .(.(y'''''''', z''''''''), nil)), nil))
G(.(x'', .(.(nil, nil), nil))) -> G(.(.(x'', .(nil, nil)), nil))
G(.(x'''', .(y0', .(.(y'''''', z''''''), nil)))) -> G(.(.(x'''', y0'), .(.(y'''''', z''''''), nil)))
G(.(.(x'''''', .(y'''''', z'''''')), .(nil, nil))) -> G(.(.(.(x'''''', .(y'''''', z'''''')), nil), nil))
G(.(.(x'''', .(nil, nil)), nil)) -> G(.(x'''', .(nil, nil)))
G(.(.(.(x'''', .(y'''', z'''')), nil), nil)) -> G(.(.(x'''', .(y'''', z'''')), nil))
G(.(.(.(x'''', nil), nil), nil)) -> G(.(.(x'''', nil), nil))
G(.(.(x'''''', nil), .(nil, nil))) -> G(.(.(.(x'''''', nil), nil), nil))
G(.(x'''', .(y0', .(nil, nil)))) -> G(.(.(x'''', y0'), .(nil, nil)))
G(.(x'''', .(y0'', .(y''0, .(y'''', z''''))))) -> G(.(.(x'''', y0''), .(y''0, .(y'''', z''''))))
G(.(.(x'''', .(y''0, .(y'''', z''''))), nil)) -> G(.(x'''', .(y''0, .(y'''', z''''))))
G(.(x'', .(.(y''''0, .(y'''''', z'''''')), nil))) -> G(.(.(x'', .(y''''0, .(y'''''', z''''''))), nil))
G(.(.(x'''', .(.(y'''''', z''''''), nil)), nil)) -> G(.(x'''', .(.(y'''''', z''''''), nil)))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

G(.(.(x'''', .(nil, nil)), nil)) -> G(.(x'''', .(nil, nil)))
G(.(.(.(x'''', .(y'''', z'''')), nil), nil)) -> G(.(.(x'''', .(y'''', z'''')), nil))
G(.(.(.(x'''', nil), nil), nil)) -> G(.(.(x'''', nil), nil))
G(.(.(x'''', .(y''0, .(y'''', z''''))), nil)) -> G(.(x'''', .(y''0, .(y'''', z''''))))
G(.(.(x'''', .(.(y'''''', z''''''), nil)), nil)) -> G(.(x'''', .(.(y'''''', z''''''), nil)))

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(G(x₁)) = 1 + x₁

POL(nil) = 0

POL(.(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 20

                 ↳Dependency Graph

Dependency Pairs:

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 21

                 ↳Polynomial Ordering

Dependency Pair:

G(.(x'''', .(y0'', .(y''0, .(y'''', z''''))))) -> G(.(.(x'''', y0''), .(y''0, .(y'''', z''''))))

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

G(.(x'''', .(y0'', .(y''0, .(y'''', z''''))))) -> G(.(.(x'''', y0''), .(y''0, .(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(G(x₁)) = x₁

POL(.(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 13

             ↳FwdInst

...

               →DP Problem 22

                 ↳Dependency Graph

Dependency Pair:

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes

POL(nil)	= 0
POL(.(x₁, x₂))	= 1 + x₁ + x₂
POL(F(x₁))	= 1 + x₁

POL(G(x₁))	= 1 + x₁
POL(nil)	= 0
POL(.(x₁, x₂))	= 1 + x₁ + x₂