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