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))
ack(s(x), y) -> f(x, x)
f(s(x), y) -> f(x, s(x))
f(x, s(y)) -> f(y, x)
f(x, y) -> ack(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)
ACK(s(x), y) -> F(x, x)
F(s(x), y) -> F(x, s(x))
F(x, s(y)) -> F(y, x)
F(x, y) -> ACK(x, y)

Furthermore, R contains one SCC.

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

The following remains to be proven:
Dependency Pairs:

F(x, y) -> ACK(x, y)
F(x, s(y)) -> F(y, x)
F(s(x), y) -> F(x, s(x))
ACK(s(x), y) -> F(x, x)
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))
ack(s(x), y) -> f(x, x)
f(s(x), y) -> f(x, s(x))
f(x, s(y)) -> f(y, x)
f(x, y) -> ack(x, y)

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