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

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

Furthermore, R contains three SCCs.

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

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS(s(x), s(y)) -> MINUS(x, 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)) =  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`
`         ↳Remaining`

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

LE(s(x), s(y)) -> LE(x, 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)) =  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`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Remaining`

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

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

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes