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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.

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

Dependency Pairs:

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

Rules:

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

The following dependency pairs can be strictly oriented:

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

The following rules can be oriented:

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

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
ACK > ack > s

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

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

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

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