We consider the TRS containing the following rules:
f(x) | → | x | (1) |
f(x) | → | f(f(x)) | (2) |
The underlying signature is as follows:
{f/1}To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:
f(x) | → | x | (1) |
All redundant rules that were added or removed can be simulated in 4 steps .
Confluence is proven using the following terminating critical-pair-closing-system R:
There are no rules.
There are no rules in the TRS. Hence, it is terminating.