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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

PLUS(s(s(x)), y) -> PLUS(x, s(y))
PLUS(x, s(s(y))) -> PLUS(s(x), y)
ACK(s(x), 0) -> ACK(x, s(0))
ACK(s(x), s(y)) -> ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), s(y)) -> PLUS(y, ack(s(x), y))
ACK(s(x), s(y)) -> ACK(s(x), y)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳AFS`

Dependency Pairs:

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

Rules:

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

The following dependency pairs can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(PLUS(x1, x2)) =  1 + x1 + x2 POL(s(x1)) =  1 + x1

resulting in one new DP problem.
Used Argument Filtering System:
PLUS(x1, x2) -> PLUS(x1, x2)
s(x1) -> s(x1)

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

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Argument Filtering and Ordering`

Dependency Pairs:

ACK(s(x), s(y)) -> ACK(s(x), y)
ACK(s(x), s(y)) -> ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), 0) -> ACK(x, s(0))

Rules:

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

The following dependency pairs can be strictly oriented:

ACK(s(x), s(y)) -> ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), 0) -> ACK(x, s(0))

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(s(x1)) =  1 + x1

resulting in one new DP problem.
Used Argument Filtering System:
ACK(x1, x2) -> x1
s(x1) -> s(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`           →DP Problem 4`
`             ↳Argument Filtering and Ordering`

Dependency Pair:

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

Rules:

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

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(ACK(x1, x2)) =  x1 + x2 POL(s(x1)) =  1 + x1

resulting in one new DP problem.
Used Argument Filtering System:
ACK(x1, x2) -> ACK(x1, x2)
s(x1) -> s(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`           →DP Problem 4`
`             ↳AFS`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

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