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))

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

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

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))

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(nil) = 1

POL(.(x₁, x₂)) = x₁ + x₂

POL(F(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Instantiation Transformation

       →DP Problem 2

         ↳Polo

Dependency Pair:

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))

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Inst

...

               →DP Problem 4

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳Polo

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Inst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

Dependency Pair:

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

The following dependency pair can be strictly oriented:

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

There are no usable rules 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

         ↳Polo

           →DP Problem 3

             ↳Inst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

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))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

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))

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(G(x₁)) = x₁

POL(nil) = 1

POL(.(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Instantiation Transformation

Dependency Pair:

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))

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Inst

...

               →DP Problem 8

                 ↳Instantiation Transformation

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Inst

...

               →DP Problem 9

                 ↳Polynomial Ordering

Dependency Pair:

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

The following dependency pair can be strictly oriented:

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

There are no usable rules 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

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Inst

...

               →DP Problem 10

                 ↳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))

Using the Dependency Graph resulted in no new DP problems.

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

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

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