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:
| s(x) | → | f(p(x)) | (5) |
| r(x) | → | s(p(x)) | (4) |
| q(x) | → | s(p(x)) | (3) |
| p(x) | → | r(x) | (2) |
| p(x) | → | q(x) | (1) |
| s(x) | → | f(r(x)) | (6) |
| s(x) | → | f(q(x)) | (7) |
| r(x) | → | f(p(p(x))) | (8) |
| r(x) | → | s(r(x)) | (9) |
| r(x) | → | s(q(x)) | (10) |
| q(x) | → | f(p(p(x))) | (11) |
| q(x) | → | s(r(x)) | (12) |
| q(x) | → | s(q(x)) | (13) |
| p(x) | → | s(p(x)) | (14) |
All redundant rules that were added or removed can be simulated in 2 steps .