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