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)

Innermost 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)


Strategy:

innermost



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