Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z]
f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(X) -> F(g(X))
F(X) -> G(X)
G(s(X)) -> G(X)
SEL(s(X), cons(Y, Z)) -> SEL(X, Z)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

Dependency Pair:

G(s(X)) -> G(X)

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Strategy:

innermost

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

G(s(X)) -> G(X)

one new Dependency Pair is created:

G(s(s(X''))) -> G(s(X''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

Dependency Pair:

G(s(s(X''))) -> G(s(X''))

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Strategy:

innermost

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

G(s(s(X''))) -> G(s(X''))

one new Dependency Pair is created:

G(s(s(s(X'''')))) -> G(s(s(X'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

Dependency Pair:

G(s(s(s(X'''')))) -> G(s(s(X'''')))

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Strategy:

innermost

The following dependency pair can be strictly oriented:

G(s(s(s(X'''')))) -> G(s(s(X'''')))

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

G(x₁) -> G(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

Dependency Pair:

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pair:

SEL(s(X), cons(Y, Z)) -> SEL(X, Z)

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Strategy:

innermost

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

SEL(s(X), cons(Y, Z)) -> SEL(X, Z)

one new Dependency Pair is created:

SEL(s(s(X'')), cons(Y, cons(Y'', Z''))) -> SEL(s(X''), cons(Y'', Z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pair:

SEL(s(s(X'')), cons(Y, cons(Y'', Z''))) -> SEL(s(X''), cons(Y'', Z''))

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Strategy:

innermost

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

SEL(s(s(X'')), cons(Y, cons(Y'', Z''))) -> SEL(s(X''), cons(Y'', Z''))

one new Dependency Pair is created:

SEL(s(s(s(X''''))), cons(Y, cons(Y''0, cons(Y'''', Z'''')))) -> SEL(s(s(X'''')), cons(Y''0, cons(Y'''', Z'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Remaining

Dependency Pair:

SEL(s(s(s(X''''))), cons(Y, cons(Y''0, cons(Y'''', Z'''')))) -> SEL(s(s(X'''')), cons(Y''0, cons(Y'''', Z'''')))

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Strategy:

innermost

The following dependency pair can be strictly oriented:

SEL(s(s(s(X''''))), cons(Y, cons(Y''0, cons(Y'''', Z'''')))) -> SEL(s(s(X'''')), cons(Y''0, cons(Y'''', Z'''')))

SEL(x₁, x₂) -> SEL(x₁, x₂)
s(x₁) -> s(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Remaining

Dependency Pair:

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

F(X) -> F(g(X))

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes