YES
by csi
The rewrite relation of the following TRS is considered.
f(x) | → | f(f(x)) |
g(x) | → | f(x) |
f(x) | → | g(x) |
To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:
f(x) | → | g(x) |
g(x) | → | f(x) |
f(x) | → | f(f(x)) |
f(x) | → | f(x) |
g(x) | → | g(x) |
g(x) | → | f(f(x)) |
f(x) | → | f(g(x)) |
f(x) | → | g(f(x)) |
f(x) | → | f(f(f(x))) |
All redundant rules that were added or removed can be simulated in 2 steps .
The duplicating rules (R) terminate relative to the other rules (S).
There are no rules in the TRS R. Hence, R/S is relative terminating.
csi