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

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

Furthermore, R contains two SCCs.

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

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

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

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

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

Dependency Pair:

Rules:

times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(0, x) -> x
plus(x, s(y)) -> s(plus(x, 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`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

TIMES(x, s(y)) -> TIMES(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(TIMES(x1, x2)) =  x2 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`

Dependency Pair:

Rules:

times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(0, x) -> x
plus(x, s(y)) -> s(plus(x, 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:
0:00 minutes