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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

INTLIST(.(x, y)) -> INTLIST(y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

INTLIST(.(x, .(x'', y''))) -> INTLIST(.(x'', y''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 4`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳FwdInst`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(INT_LIST(x1)) =  1 + x1 POL(.(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳FwdInst`

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Forward Instantiation Transformation`

Dependency Pairs:

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

Rules:

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

`   R`
`     ↳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:

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

`   R`
`     ↳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:

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

`   R`
`     ↳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:

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 11`
`                 ↳Polynomial 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:

int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(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 w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(s(x1)) =  1 + x1 POL(INT(x1, x2)) =  x2

resulting in one new DP problem.

`   R`
`     ↳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:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳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:

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳FwdInst`
`           →DP Problem 7`
`             ↳FwdInst`
`             ...`
`               →DP Problem 9`
`                 ↳Polynomial Ordering`

Dependency Pair:

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

Rules:

int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> intlist(int(x, y))
intlist(nil) -> nil
intlist(.(x, y)) -> .(s(x), intlist(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 w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(s(x1)) =  1 + x1 POL(INT(x1, x2)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳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:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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