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