Term Rewriting System R:
[y, z, x]
quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
QUOT(x, 0, s(z)) -> PLUS(z, s(0))
PLUS(s(x), y) -> PLUS(x, y)

Furthermore, R contains two SCCs.

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

Dependency Pair:

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

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

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

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

Dependency Pair:

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

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 4`
`                 ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

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

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(PLUS(x1, x2)) =  1 + x1 + x2 POL(s(x1)) =  1 + x1

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

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

Dependency Pair:

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

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:

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
two new Dependency Pairs are created:

QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))
QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))
QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

QUOT(s(x), s(y), z) -> QUOT(x, y, z)
three new Dependency Pairs are created:

QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
QUOT(s(x''), s(0), s(0)) -> QUOT(x'', 0, s(0))
QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))

The transformation is resulting in two new DP problems:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))
QUOT(s(x''), s(0), s(0)) -> QUOT(x'', 0, s(0))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

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

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 9`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))
QUOT(s(x''), s(0), s(0)) -> QUOT(x'', 0, s(0))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

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

QUOT(s(s(x'''''')), s(0), s(0)) -> QUOT(s(x''''''), 0, s(0))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 11`
`                 ↳Argument Filtering and Ordering`

Dependency Pairs:

QUOT(s(s(x'''''')), s(0), s(0)) -> QUOT(s(x''''''), 0, s(0))
QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

QUOT(s(s(x'''''')), s(0), s(0)) -> QUOT(s(x''''''), 0, s(0))

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(s(x1)) =  1 + x1

resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2, x3) -> x1
s(x1) -> s(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 14`
`                 ↳Dependency Graph`

Dependency Pair:

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pairs:

QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))
QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))
two new Dependency Pairs are created:

QUOT(x, 0, s(s(0))) -> QUOT(x, s(s(0)), s(s(0)))
QUOT(x, 0, s(s(s(x''')))) -> QUOT(x, s(s(plus(x''', s(0)))), s(s(s(x'''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 10`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

QUOT(x, 0, s(s(s(x''')))) -> QUOT(x, s(s(plus(x''', s(0)))), s(s(s(x'''))))
QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
QUOT(x, 0, s(s(0))) -> QUOT(x, s(s(0)), s(s(0)))
QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
two new Dependency Pairs are created:

QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 12`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

QUOT(x, 0, s(s(0))) -> QUOT(x, s(s(0)), s(s(0)))
QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))
QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))
QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
QUOT(x, 0, s(s(s(x''')))) -> QUOT(x, s(s(plus(x''', s(0)))), s(s(s(x'''))))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))
two new Dependency Pairs are created:

QUOT(s(x'''), s(0), s(s(0))) -> QUOT(x''', 0, s(s(0)))
QUOT(s(x'''), s(0), s(s(s(x'''''')))) -> QUOT(x''', 0, s(s(s(x''''''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 13`
`                 ↳Argument Filtering and Ordering`

Dependency Pairs:

QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
QUOT(x, 0, s(s(s(x''')))) -> QUOT(x, s(s(plus(x''', s(0)))), s(s(s(x'''))))
QUOT(s(x'''), s(0), s(s(s(x'''''')))) -> QUOT(x''', 0, s(s(s(x''''''))))
QUOT(s(x'''), s(0), s(s(0))) -> QUOT(x''', 0, s(s(0)))
QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))
QUOT(x, 0, s(s(0))) -> QUOT(x, s(s(0)), s(s(0)))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
QUOT(s(x'''), s(0), s(s(s(x'''''')))) -> QUOT(x''', 0, s(s(s(x''''''))))
QUOT(s(x'''), s(0), s(s(0))) -> QUOT(x''', 0, s(s(0)))
QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))

The following usable rules for innermost can be oriented:

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

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

resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2, x3) -> x1
s(x1) -> s(x1)
plus(x1, x2) -> plus(x1, x2)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 15`
`                 ↳Dependency Graph`

Dependency Pairs:

QUOT(x, 0, s(s(s(x''')))) -> QUOT(x, s(s(plus(x''', s(0)))), s(s(s(x'''))))
QUOT(x, 0, s(s(0))) -> QUOT(x, s(s(0)), s(s(0)))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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