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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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

Furthermore, R contains three SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

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

Rules:

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pairs:

MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
MINUS(s(x), s(y)) -> MINUS(x, y)

Rules:

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

The following dependency pair can be strictly oriented:

MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))

Additionally, the following rules can be oriented:

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

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

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

Rules:

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Polo`
`             ...`
`               →DP Problem 6`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polynomial Ordering`

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 7`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

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