Termination Proof using AProVE

Term Rewriting System R:
[l, x, k, a, b, c]
f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(cons(x, k), l) -> G(k, l, cons(x, k))
G(a, b, c) -> F(a, cons(b, c))

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Instantiation Transformation

Dependency Pairs:

G(a, b, c) -> F(a, cons(b, c))
F(cons(x, k), l) -> G(k, l, cons(x, k))

Rules:

f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))

Strategy:

innermost

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

F(cons(x, k), l) -> G(k, l, cons(x, k))

one new Dependency Pair is created:

F(cons(x', k'), cons(b'', c'')) -> G(k', cons(b'', c''), cons(x', k'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Instantiation Transformation

Dependency Pairs:

F(cons(x', k'), cons(b'', c'')) -> G(k', cons(b'', c''), cons(x', k'))
G(a, b, c) -> F(a, cons(b, c))

Rules:

f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))

Strategy:

innermost

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

G(a, b, c) -> F(a, cons(b, c))

one new Dependency Pair is created:

G(a', cons(b'''', c''''), cons(x''', a')) -> F(a', cons(cons(b'''', c''''), cons(x''', a')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Inst

...

               →DP Problem 3

                 ↳Instantiation Transformation

Dependency Pairs:

G(a', cons(b'''', c''''), cons(x''', a')) -> F(a', cons(cons(b'''', c''''), cons(x''', a')))
F(cons(x', k'), cons(b'', c'')) -> G(k', cons(b'', c''), cons(x', k'))

Rules:

f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))

Strategy:

innermost

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

F(cons(x', k'), cons(b'', c'')) -> G(k', cons(b'', c''), cons(x', k'))

one new Dependency Pair is created:

F(cons(x'', k''), cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k'')))) -> G(k'', cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k''))), cons(x'', k''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Inst

...

               →DP Problem 4

                 ↳Instantiation Transformation

Dependency Pairs:

F(cons(x'', k''), cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k'')))) -> G(k'', cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k''))), cons(x'', k''))
G(a', cons(b'''', c''''), cons(x''', a')) -> F(a', cons(cons(b'''', c''''), cons(x''', a')))

Rules:

f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))

Strategy:

innermost

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

G(a', cons(b'''', c''''), cons(x''', a')) -> F(a', cons(cons(b'''', c''''), cons(x''', a')))

one new Dependency Pair is created:

G(a'', cons(cons(b'''''''', c''''''''), cons(x''''''', cons(x''''', a''))), cons(x''''', a'')) -> F(a'', cons(cons(cons(b'''''''', c''''''''), cons(x''''''', cons(x''''', a''))), cons(x''''', a'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Inst

...

               →DP Problem 5

                 ↳Argument Filtering and Ordering

Dependency Pairs:

G(a'', cons(cons(b'''''''', c''''''''), cons(x''''''', cons(x''''', a''))), cons(x''''', a'')) -> F(a'', cons(cons(cons(b'''''''', c''''''''), cons(x''''''', cons(x''''', a''))), cons(x''''', a'')))
F(cons(x'', k''), cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k'')))) -> G(k'', cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k''))), cons(x'', k''))

Rules:

f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(cons(x'', k''), cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k'')))) -> G(k'', cons(cons(b'''''', c''''''), cons(x''''', cons(x'', k''))), cons(x'', k''))

There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂) -> x₁
cons(x₁, x₂) -> cons(x₁, x₂)
G(x₁, x₂, x₃) -> x₁

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 2

             ↳Inst

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pair:

Rules:

f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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