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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
+'(s(x), y) -> +'(x, y)
*'(x, s(y)) -> +'(x, *(x, y))
*'(x, s(y)) -> *'(x, y)
F(s(x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))
F(s(x)) -> -'(*(s(s(0)), s(x)), s(s(x)))
F(s(x)) -> *'(s(s(0)), s(x))

Furthermore, R contains four SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.
Used Argument Filtering System:
-'(x1, x2) -> -'(x1, x2)
s(x1) -> s(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.
Used Argument Filtering System:
+'(x1, x2) -> +'(x1, x2)
s(x1) -> s(x1)

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

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 4`
`         ↳Rw`

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.
Used Argument Filtering System:
*'(x1, x2) -> *'(x1, x2)
s(x1) -> s(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`           →DP Problem 7`
`             ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Rw`

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rewriting Transformation`

Dependency Pair:

F(s(x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))
one new Dependency Pair is created:

F(s(x)) -> F(-(+(s(s(0)), *(s(s(0)), x)), s(s(x))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rewriting Transformation`

Dependency Pair:

F(s(x)) -> F(-(+(s(s(0)), *(s(s(0)), x)), s(s(x))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(x)) -> F(-(+(s(s(0)), *(s(s(0)), x)), s(s(x))))
one new Dependency Pair is created:

F(s(x)) -> F(-(s(+(s(0), *(s(s(0)), x))), s(s(x))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 9`
`                 ↳Rewriting Transformation`

Dependency Pair:

F(s(x)) -> F(-(s(+(s(0), *(s(s(0)), x))), s(s(x))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(x)) -> F(-(s(+(s(0), *(s(s(0)), x))), s(s(x))))
one new Dependency Pair is created:

F(s(x)) -> F(-(+(s(0), *(s(s(0)), x)), s(x)))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 10`
`                 ↳Rewriting Transformation`

Dependency Pair:

F(s(x)) -> F(-(+(s(0), *(s(s(0)), x)), s(x)))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(x)) -> F(-(+(s(0), *(s(s(0)), x)), s(x)))
one new Dependency Pair is created:

F(s(x)) -> F(-(s(+(0, *(s(s(0)), x))), s(x)))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 11`
`                 ↳Rewriting Transformation`

Dependency Pair:

F(s(x)) -> F(-(s(+(0, *(s(s(0)), x))), s(x)))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(x)) -> F(-(s(+(0, *(s(s(0)), x))), s(x)))
one new Dependency Pair is created:

F(s(x)) -> F(-(+(0, *(s(s(0)), x)), x))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 12`
`                 ↳Rewriting Transformation`

Dependency Pair:

F(s(x)) -> F(-(+(0, *(s(s(0)), x)), x))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(x)) -> F(-(+(0, *(s(s(0)), x)), x))
one new Dependency Pair is created:

F(s(x)) -> F(-(*(s(s(0)), x), x))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 13`
`                 ↳Narrowing Transformation`

Dependency Pair:

F(s(x)) -> F(-(*(s(s(0)), x), x))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(x)) -> F(-(*(s(s(0)), x), x))
three new Dependency Pairs are created:

F(s(0)) -> F(*(s(s(0)), 0))
F(s(0)) -> F(-(0, 0))
F(s(s(y'))) -> F(-(+(s(s(0)), *(s(s(0)), y')), s(y')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 14`
`                 ↳Rewriting Transformation`

Dependency Pairs:

F(s(s(y'))) -> F(-(+(s(s(0)), *(s(s(0)), y')), s(y')))
F(s(0)) -> F(-(0, 0))
F(s(0)) -> F(*(s(s(0)), 0))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(0)) -> F(*(s(s(0)), 0))
one new Dependency Pair is created:

F(s(0)) -> F(0)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 15`
`                 ↳Rewriting Transformation`

Dependency Pairs:

F(s(0)) -> F(-(0, 0))
F(s(s(y'))) -> F(-(+(s(s(0)), *(s(s(0)), y')), s(y')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(0)) -> F(-(0, 0))
one new Dependency Pair is created:

F(s(0)) -> F(0)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 16`
`                 ↳Rewriting Transformation`

Dependency Pair:

F(s(s(y'))) -> F(-(+(s(s(0)), *(s(s(0)), y')), s(y')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(s(s(0)), *(s(s(0)), y')), s(y')))
one new Dependency Pair is created:

F(s(s(y'))) -> F(-(s(+(s(0), *(s(s(0)), y'))), s(y')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 17`
`                 ↳Rewriting Transformation`

Dependency Pair:

F(s(s(y'))) -> F(-(s(+(s(0), *(s(s(0)), y'))), s(y')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(s(y'))) -> F(-(s(+(s(0), *(s(s(0)), y'))), s(y')))
one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(s(0), *(s(s(0)), y')), y'))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 18`
`                 ↳Rewriting Transformation`

Dependency Pair:

F(s(s(y'))) -> F(-(+(s(0), *(s(s(0)), y')), y'))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(s(y'))) -> F(-(+(s(0), *(s(s(0)), y')), y'))
one new Dependency Pair is created:

F(s(s(y'))) -> F(-(s(+(0, *(s(s(0)), y'))), y'))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 19`
`                 ↳Rewriting Transformation`

Dependency Pair:

F(s(s(y'))) -> F(-(s(+(0, *(s(s(0)), y'))), y'))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

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

F(s(s(y'))) -> F(-(s(+(0, *(s(s(0)), y'))), y'))
one new Dependency Pair is created:

F(s(s(y'))) -> F(-(s(*(s(s(0)), y')), y'))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Rw`
`           →DP Problem 8`
`             ↳Rw`
`             ...`
`               →DP Problem 20`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pair:

F(s(s(y'))) -> F(-(s(*(s(s(0)), y')), y'))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:01 minutes