We consider the TRS containing the following rules:
| f(g(x)) | → | h(x,x) | (1) |
| g(a) | → | b | (2) |
| f(x) | → | h(x,x) | (3) |
| b | → | a | (4) |
| h(x,y) | → | h(g(x),g(y)) | (5) |
| g(x) | → | x | (6) |
| a | → | b | (7) |
The underlying signature is as follows:
{f/1, g/1, h/2, a/0, b/0}To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:
| a | → | b | (7) |
| g(x) | → | x | (6) |
| h(x,y) | → | h(g(x),g(y)) | (5) |
| b | → | a | (4) |
| f(x) | → | h(x,x) | (3) |
| g(a) | → | b | (2) |
| f(g(x)) | → | h(x,x) | (1) |
| a | → | a | (8) |
| h(x,y) | → | h(g(x),y) | (9) |
| h(x,y) | → | h(x,g(y)) | (10) |
| h(x,y) | → | h(g(g(x)),g(g(y))) | (11) |
| b | → | b | (12) |
| f(x) | → | h(g(x),g(x)) | (13) |
| g(a) | → | a | (14) |
| f(g(x)) | → | h(g(x),g(x)) | (15) |
All redundant rules that were added or removed can be simulated in 2 steps .