Term Rewriting System R:
[x, y]
p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

FACT(s(x)) -> *'(s(x), fact(p(s(x))))
FACT(s(x)) -> FACT(p(s(x)))
FACT(s(x)) -> P(s(x))
*'(s(x), y) -> +'(*(x, y), y)
*'(s(x), y) -> *'(x, y)
+'(x, s(y)) -> +'(x, y)

Furthermore, R contains three SCCs.

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

Dependency Pair:

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

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

+'(x, s(y)) -> +'(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 4`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳Rw`

Dependency Pair:

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

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

+'(x'', s(s(y''))) -> +'(x'', s(y''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 4`
`             ↳FwdInst`
`             ...`
`               →DP Problem 5`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳Rw`

Dependency Pair:

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

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

+'(x'''', s(s(s(y'''')))) -> +'(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(+'(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 4`
`             ↳FwdInst`
`             ...`
`               →DP Problem 6`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳Rw`

Dependency Pair:

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, 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`
`       →DP Problem 3`
`         ↳Rw`

Dependency Pair:

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

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

*'(s(x), y) -> *'(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 7`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳Rw`

Dependency Pair:

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

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

*'(s(s(x'')), y'') -> *'(s(x''), y'')
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 7`
`             ↳FwdInst`
`             ...`
`               →DP Problem 8`
`                 ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Rw`

Dependency Pair:

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

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

*'(s(s(s(x''''))), y'''') -> *'(s(s(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(*'(x1, x2)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 7`
`             ↳FwdInst`
`             ...`
`               →DP Problem 9`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Rw`

Dependency Pair:

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, 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 3`
`         ↳Rewriting Transformation`

Dependency Pair:

FACT(s(x)) -> FACT(p(s(x)))

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

FACT(s(x)) -> FACT(p(s(x)))
one new Dependency Pair is created:

FACT(s(x)) -> FACT(x)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Forward Instantiation Transformation`

Dependency Pair:

FACT(s(x)) -> FACT(x)

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

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

FACT(s(x)) -> FACT(x)
one new Dependency Pair is created:

FACT(s(s(x''))) -> FACT(s(x''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳Rw`
`           →DP Problem 10`
`             ↳FwdInst`
`             ...`
`               →DP Problem 11`
`                 ↳Polynomial Ordering`

Dependency Pair:

FACT(s(s(x''))) -> FACT(s(x''))

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

FACT(s(s(x''))) -> FACT(s(x''))

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(FACT(x1)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳Rw`
`           →DP Problem 10`
`             ↳FwdInst`
`             ...`
`               →DP Problem 12`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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