Term Rewriting System R:
[Y, X]
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

PLUS(s(X), Y) -> PLUS(X, Y)
MIN(s(X), s(Y)) -> MIN(X, Y)
MIN(min(X, Y), Z) -> MIN(X, plus(Y, Z))
MIN(min(X, Y), Z) -> PLUS(Y, Z)
QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MIN(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:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

The following dependency pair can be strictly oriented:

PLUS(s(X), Y) -> PLUS(X, Y)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(PLUS(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:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(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:

MIN(min(X, Y), Z) -> MIN(X, plus(Y, Z))
MIN(s(X), s(Y)) -> MIN(X, Y)

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

The following dependency pair can be strictly oriented:

MIN(min(X, Y), Z) -> MIN(X, plus(Y, Z))

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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

MIN(s(X), s(Y)) -> MIN(X, Y)

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

The following dependency pair can be strictly oriented:

MIN(s(X), s(Y)) -> MIN(X, Y)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(MIN(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:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(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(min(X, Y), s(Y))

Rules:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

The following dependency pair can be strictly oriented:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

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

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(Z) =  0 POL(plus(x1, x2)) =  0 POL(QUOT(x1, x2)) =  x1 POL(0) =  1 POL(min(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:

plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
min(min(X, Y), Z) -> min(X, plus(Y, Z))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))

Using the Dependency Graph resulted in no new DP problems.

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