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