We consider the TRS containing the following rules:
f(b) | → | c | (1) |
c | → | c | (2) |
f(c) | → | b | (3) |
a | → | b | (4) |
The underlying signature is as follows:
{f/1, b/0, c/0, a/0}To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:
a | → | b | (4) |
f(c) | → | b | (3) |
f(b) | → | c | (1) |
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:
a | → | b | (4) |
f(c) | → | b | (3) |
f(b) | → | c | (1) |
All redundant rules that were added or removed can be simulated in 2 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.