We consider the TRS containing the following rules:
| +(x,0) | → | x | (1) |
| +(x,s(y)) | → | s(+(x,y)) | (2) |
| d(0) | → | 0 | (3) |
| d(s(x)) | → | s(s(d(x))) | (4) |
| f(0) | → | 0 | (5) |
| f(s(x)) | → | +(+(s(x),s(x)),s(x)) | (6) |
| f(g(0)) | → | +(+(g(0),g(0)),g(0)) | (7) |
| g(x) | → | s(d(x)) | (8) |
The underlying signature is as follows:
{+/2, 0/0, s/1, d/1, f/1, g/1}To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:
| +(x,0) | → | x | (1) |
| +(x,s(y)) | → | s(+(x,y)) | (2) |
| d(0) | → | 0 | (3) |
| d(s(x)) | → | s(s(d(x))) | (4) |
| f(0) | → | 0 | (5) |
| f(s(x)) | → | +(+(s(x),s(x)),s(x)) | (6) |
| g(x) | → | s(d(x)) | (8) |
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.