Termination Proof using AProVE

Term Rewriting System R:
[x, y]
intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

INTLIST(cons(x, y)) -> INTLIST(y)
INT(s(x), s(y)) -> INTLIST(int(x, y))
INT(s(x), s(y)) -> INT(x, y)
INT(0, s(y)) -> INT(s(0), s(y))

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

INTLIST(cons(x, y)) -> INTLIST(y)

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

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

INTLIST(cons(x, y)) -> INTLIST(y)

one new Dependency Pair is created:

INTLIST(cons(x, cons(x'', y''))) -> INTLIST(cons(x'', y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

INTLIST(cons(x, cons(x'', y''))) -> INTLIST(cons(x'', y''))

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

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

INTLIST(cons(x, cons(x'', y''))) -> INTLIST(cons(x'', y''))

one new Dependency Pair is created:

INTLIST(cons(x, cons(x'''', cons(x''''', y'''')))) -> INTLIST(cons(x'''', cons(x''''', y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

INTLIST(cons(x, cons(x'''', cons(x''''', y'''')))) -> INTLIST(cons(x'''', cons(x''''', y'''')))

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

INTLIST(cons(x, cons(x'''', cons(x''''', y'''')))) -> INTLIST(cons(x'''', cons(x''''', y'''')))

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

INTLIST(x₁) -> INTLIST(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

Dependency Pairs:

INT(0, s(y)) -> INT(s(0), s(y))
INT(s(x), s(y)) -> INT(x, y)

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

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

INT(s(x), s(y)) -> INT(x, y)

two new Dependency Pairs are created:

INT(s(s(x'')), s(s(y''))) -> INT(s(x''), s(y''))
INT(s(0), s(s(y''))) -> INT(0, s(y''))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳Forward Instantiation Transformation

           →DP Problem 7

             ↳FwdInst

Dependency Pairs:

INT(s(0), s(s(y''))) -> INT(0, s(y''))
INT(0, s(y)) -> INT(s(0), s(y))

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

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

INT(0, s(y)) -> INT(s(0), s(y))

one new Dependency Pair is created:

INT(0, s(s(y''''))) -> INT(s(0), s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Forward Instantiation Transformation

           →DP Problem 7

             ↳FwdInst

Dependency Pairs:

INT(0, s(s(y''''))) -> INT(s(0), s(s(y'''')))
INT(s(0), s(s(y''))) -> INT(0, s(y''))

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

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

INT(s(0), s(s(y''))) -> INT(0, s(y''))

one new Dependency Pair is created:

INT(s(0), s(s(s(y'''''')))) -> INT(0, s(s(y'''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 10

                 ↳Forward Instantiation Transformation

           →DP Problem 7

             ↳FwdInst

Dependency Pairs:

INT(s(0), s(s(s(y'''''')))) -> INT(0, s(s(y'''''')))
INT(0, s(s(y''''))) -> INT(s(0), s(s(y'''')))

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

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

INT(0, s(s(y''''))) -> INT(s(0), s(s(y'''')))

one new Dependency Pair is created:

INT(0, s(s(s(y'''''''')))) -> INT(s(0), s(s(s(y''''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 11

                 ↳Argument Filtering and Ordering

           →DP Problem 7

             ↳FwdInst

Dependency Pairs:

INT(0, s(s(s(y'''''''')))) -> INT(s(0), s(s(s(y''''''''))))
INT(s(0), s(s(s(y'''''')))) -> INT(0, s(s(y'''''')))

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

INT(s(0), s(s(s(y'''''')))) -> INT(0, s(s(y'''''')))

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

INT(x₁, x₂) -> x₂
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 13

                 ↳Dependency Graph

           →DP Problem 7

             ↳FwdInst

Dependency Pair:

INT(0, s(s(s(y'''''''')))) -> INT(s(0), s(s(s(y''''''''))))

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

Dependency Pair:

INT(s(s(x'')), s(s(y''))) -> INT(s(x''), s(y''))

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

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

INT(s(s(x'')), s(s(y''))) -> INT(s(x''), s(y''))

one new Dependency Pair is created:

INT(s(s(s(x''''))), s(s(s(y'''')))) -> INT(s(s(x'''')), s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Argument Filtering and Ordering

Dependency Pair:

INT(s(s(s(x''''))), s(s(s(y'''')))) -> INT(s(s(x'''')), s(s(y'''')))

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

INT(s(s(s(x''''))), s(s(s(y'''')))) -> INT(s(s(x'''')), s(s(y'''')))

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

INT(x₁, x₂) -> INT(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 12

                 ↳Dependency Graph

Dependency Pair:

Rules:

intlist(nil) -> nil
intlist(cons(x, y)) -> cons(s(x), intlist(y))
intlist(cons(x, nil)) -> cons(s(x), nil)
int(s(x), 0) -> nil
int(x, x) -> cons(x, nil)
int(s(x), s(y)) -> intlist(int(x, y))
int(0, s(y)) -> cons(0, int(s(0), s(y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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