Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(s(X)) -> F(X)
G(cons(0, Y)) -> G(Y)
H(cons(X, Y)) -> H(g(cons(X, Y)))
H(cons(X, Y)) -> G(cons(X, Y))

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

F(s(X)) -> F(X)

Rules:

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

Strategy:

innermost

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

F(s(X)) -> F(X)

one new Dependency Pair is created:

F(s(s(X''))) -> F(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

         ↳Nar

Dependency Pair:

F(s(s(X''))) -> F(s(X''))

Rules:

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

Strategy:

innermost

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

F(s(s(X''))) -> F(s(X''))

one new Dependency Pair is created:

F(s(s(s(X'''')))) -> F(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

         ↳Nar

Dependency Pair:

F(s(s(s(X'''')))) -> F(s(s(X'''')))

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(s(s(s(X'''')))) -> F(s(s(X'''')))

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)) = 1 + x₁

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

F(x₁) -> F(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

         ↳Nar

Dependency Pair:

Rules:

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

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

         ↳Nar

Dependency Pair:

G(cons(0, Y)) -> G(Y)

Rules:

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

Strategy:

innermost

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

G(cons(0, Y)) -> G(Y)

one new Dependency Pair is created:

G(cons(0, cons(0, Y''))) -> G(cons(0, Y''))

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

         ↳Nar

Dependency Pair:

G(cons(0, cons(0, Y''))) -> G(cons(0, Y''))

Rules:

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

Strategy:

innermost

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

G(cons(0, cons(0, Y''))) -> G(cons(0, Y''))

one new Dependency Pair is created:

G(cons(0, cons(0, cons(0, Y'''')))) -> G(cons(0, cons(0, Y'''')))

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

         ↳Nar

Dependency Pair:

G(cons(0, cons(0, cons(0, Y'''')))) -> G(cons(0, cons(0, Y'''')))

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

G(cons(0, cons(0, cons(0, Y'''')))) -> G(cons(0, cons(0, 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(0) = 0

POL(G(x₁)) = 1 + x₁

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

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

G(x₁) -> G(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

         ↳Nar

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pair:

H(cons(X, Y)) -> H(g(cons(X, Y)))

Rules:

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

Strategy:

innermost

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

H(cons(X, Y)) -> H(g(cons(X, Y)))

two new Dependency Pairs are created:

H(cons(0, Y'')) -> H(g(Y''))
H(cons(s(X''), Y'')) -> H(s(X''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Narrowing Transformation

Dependency Pair:

H(cons(0, Y'')) -> H(g(Y''))

Rules:

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

Strategy:

innermost

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

H(cons(0, Y'')) -> H(g(Y''))

two new Dependency Pairs are created:

H(cons(0, cons(0, Y'))) -> H(g(Y'))
H(cons(0, cons(s(X'), Y'))) -> H(s(X'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 11

                 ↳Forward Instantiation Transformation

Dependency Pair:

H(cons(0, cons(0, Y'))) -> H(g(Y'))

Rules:

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

Strategy:

innermost

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

H(cons(0, cons(0, Y'))) -> H(g(Y'))

no new Dependency Pairs are created.
The transformation is resulting in no new DP problems.

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

POL(0)	= 0
POL(G(x₁))	= 1 + x₁
POL(cons(x₁, x₂))	= 1 + x₁ + x₂