We consider the TRS containing the following rules:
| f(x) | → | g(k(x)) | (1) |
| f(x) | → | a | (2) |
| g(x) | → | a | (3) |
| k(a) | → | k(k(a)) | (4) |
The underlying signature is as follows:
{f/1, g/1, k/1, a/0}To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:
| f(x) | → | a | (2) |
| g(x) | → | a | (3) |
| k(a) | → | k(k(a)) | (4) |
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.