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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
F(s(x), y) -> F(p(-(s(x), y)), p(-(y, s(x))))
F(s(x), y) -> P(-(s(x), y))
F(s(x), y) -> -'(s(x), y)
F(s(x), y) -> P(-(y, s(x)))
F(s(x), y) -> -'(y, s(x))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(x, s(y)) -> P(-(x, s(y)))
F(x, s(y)) -> -'(x, s(y))
F(x, s(y)) -> P(-(s(y), x))
F(x, s(y)) -> -'(s(y), x)

Furthermore, R contains two SCCs.

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

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

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

-'(s(s(x'')), s(s(y''))) -> -'(s(x''), s(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`
`         ↳Nar`

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

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

-'(s(s(s(x''''))), s(s(s(y'''')))) -> -'(s(s(x'''')), s(s(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`
`         ↳Nar`

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

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

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Narrowing Transformation`

Dependency Pairs:

F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x), y) -> F(p(-(s(x), y)), p(-(y, s(x))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(s(x), y) -> F(p(-(s(x), y)), p(-(y, s(x))))
three new Dependency Pairs are created:

F(s(x''), 0) -> F(p(s(x'')), p(-(0, s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rewriting Transformation`

Dependency Pairs:

F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(s(x''), 0) -> F(p(s(x'')), p(-(0, s(x''))))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

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

F(s(x''), 0) -> F(x'', p(-(0, s(x''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 7`
`                 ↳Rewriting Transformation`

Dependency Pairs:

F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 8`
`                 ↳Rewriting Transformation`

Dependency Pairs:

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(s(x0), s(x'')) -> F(p(-(s(x0), s(x''))), p(-(x'', x0)))
one new Dependency Pair is created:

F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 9`
`                 ↳Narrowing Transformation`

Dependency Pairs:

F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(x, s(y)) -> F(p(-(x, s(y))), p(-(s(y), x)))
three new Dependency Pairs are created:

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(0, s(y')) -> F(p(-(0, s(y'))), p(s(y')))
F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 10`
`                 ↳Rewriting Transformation`

Dependency Pairs:

F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))
F(0, s(y')) -> F(p(-(0, s(y'))), p(s(y')))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(s(y''), s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 11`
`                 ↳Rewriting Transformation`

Dependency Pairs:

F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(0, s(y')) -> F(p(-(0, s(y'))), p(s(y')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 12`
`                 ↳Rewriting Transformation`

Dependency Pairs:

F(0, s(y')) -> F(p(-(0, s(y'))), y')
F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(s(y''), s(y0)) -> F(p(-(s(y''), s(y0))), p(-(y0, y'')))
one new Dependency Pair is created:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 13`
`                 ↳Narrowing Transformation`

Dependency Pairs:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
F(0, s(y')) -> F(p(-(0, s(y'))), y')

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(s(x''), 0) -> F(x'', p(-(0, s(x''))))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 14`
`                 ↳Narrowing Transformation`

Dependency Pairs:

F(0, s(y')) -> F(p(-(0, s(y'))), y')
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

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

F(0, s(y')) -> F(p(-(0, s(y'))), y')
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 15`
`                 ↳Polynomial Ordering`

Dependency Pairs:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(s(y''), s(y0)) -> F(p(-(y'', y0)), p(-(y0, y'')))
F(s(x0), s(x'')) -> F(p(-(x0, x'')), p(-(x'', x0)))
F(s(x''), s(y'')) -> F(p(-(x'', y'')), p(-(y'', x'')))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

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

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Rw`
`             ...`
`               →DP Problem 16`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
p(s(x)) -> x
f(s(x), y) -> f(p(-(s(x), y)), p(-(y, s(x))))
f(x, s(y)) -> f(p(-(x, s(y))), p(-(s(y), x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes