Term Rewriting System R:
[x, y]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
LE(s(x), s(y)) -> LE(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(s(x), s(y)), s(y))
QUOT(s(x), s(y)) -> MINUS(s(x), s(y))

Furthermore, R contains three SCCs.

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

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Strategy:

innermost

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

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

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

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Strategy:

innermost

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

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

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

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(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:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Strategy:

innermost

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

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

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

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Strategy:

innermost

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

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

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(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 7`
`             ↳FwdInst`
`             ...`
`               →DP Problem 8`
`                 ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Rw`

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(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:

QUOT(s(x), s(y)) -> QUOT(minus(s(x), s(y)), s(y))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Strategy:

innermost

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

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

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

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`
`             ↳Narrowing Transformation`

Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Strategy:

innermost

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

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

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

The transformation is resulting in two new DP problems:

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

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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(QUOT(x1, x2)) =  x1 POL(0) =  0 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`
`             ↳Nar`
`             ...`
`               →DP Problem 13`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(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`
`         ↳Rw`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 12`
`                 ↳Polynomial Ordering`

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

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

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

resulting in one new DP problem.

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