We consider the TRS containing the following rules:
| p(x) | → | q(x) | (1) |
| p(x) | → | r(x) | (2) |
| q(x) | → | s(p(x)) | (3) |
| r(x) | → | s(p(x)) | (4) |
| s(x) | → | f(p(x)) | (5) |
The underlying signature is as follows:
{p/1, q/1, r/1, s/1, f/1}To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:
| p(x) | → | r(x) | (2) |
| q(x) | → | s(p(x)) | (3) |
| r(x) | → | s(p(x)) | (4) |
| s(x) | → | f(p(x)) | (5) |
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.