Termination Proof using AProVE

Term Rewriting System R:
[x, y]
g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

G(c(x, s(y))) -> G(c(s(x), y))
F(c(s(x), y)) -> F(c(x, s(y)))
F(f(x)) -> F(d(f(x)))

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Instantiation Transformation

       →DP Problem 2

         ↳Inst

Dependency Pair:

G(c(x, s(y))) -> G(c(s(x), y))

Rules:

g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x

Strategy:

innermost

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

G(c(x, s(y))) -> G(c(s(x), y))

one new Dependency Pair is created:

G(c(s(x''), s(y''))) -> G(c(s(s(x'')), y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 3

             ↳Instantiation Transformation

       →DP Problem 2

         ↳Inst

Dependency Pair:

G(c(s(x''), s(y''))) -> G(c(s(s(x'')), y''))

Rules:

g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x

Strategy:

innermost

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

G(c(s(x''), s(y''))) -> G(c(s(s(x'')), y''))

one new Dependency Pair is created:

G(c(s(s(x'''')), s(y''''))) -> G(c(s(s(s(x''''))), y''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 3

             ↳Inst

...

               →DP Problem 4

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Inst

Dependency Pair:

G(c(s(s(x'''')), s(y''''))) -> G(c(s(s(s(x''''))), y''''))

Rules:

g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x

Strategy:

innermost

The following dependency pair can be strictly oriented:

G(c(s(s(x'''')), s(y''''))) -> G(c(s(s(s(x''''))), y''''))

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(G(x₁)) = x₁

POL(s(x₁)) = 1 + x₁

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

G(x₁) -> G(x₁)
c(x₁, x₂) -> x₂
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 3

             ↳Inst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Inst

Dependency Pair:

Rules:

g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Instantiation Transformation

Dependency Pair:

F(c(s(x), y)) -> F(c(x, s(y)))

Rules:

g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x

Strategy:

innermost

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

F(c(s(x), y)) -> F(c(x, s(y)))

one new Dependency Pair is created:

F(c(s(x''), s(y''))) -> F(c(x'', s(s(y''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Inst

           →DP Problem 6

             ↳Instantiation Transformation

Dependency Pair:

F(c(s(x''), s(y''))) -> F(c(x'', s(s(y''))))

Rules:

g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x

Strategy:

innermost

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

F(c(s(x''), s(y''))) -> F(c(x'', s(s(y''))))

one new Dependency Pair is created:

F(c(s(x''''), s(s(y'''')))) -> F(c(x'''', s(s(s(y'''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Inst

           →DP Problem 6

             ↳Inst

...

               →DP Problem 7

                 ↳Argument Filtering and Ordering

Dependency Pair:

F(c(s(x''''), s(s(y'''')))) -> F(c(x'''', s(s(s(y'''')))))

Rules:

g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(c(s(x''''), s(s(y'''')))) -> F(c(x'''', s(s(s(y'''')))))

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(F(x₁)) = x₁

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

F(x₁) -> F(x₁)
c(x₁, x₂) -> x₁
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Inst

           →DP Problem 6

             ↳Inst

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

g(c(x, s(y))) -> g(c(s(x), y))
f(c(s(x), y)) -> f(c(x, s(y)))
f(f(x)) -> f(d(f(x)))
f(x) -> x

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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