Certification Problem
Input (COPS 1080)
We consider two TRSs R and S where R contains the rules
|
f(x) |
→ |
f(f(x)) |
(1) |
|
g(x) |
→ |
f(x) |
(2) |
|
f(x) |
→ |
g(x) |
(3) |
and S contains the following rules:
|
f(c) |
→ |
g(c) |
(4) |
|
g(c) |
→ |
f(c) |
(5) |
| c |
→ |
d |
(6) |
The underlying signature is as follows:
{c/0, d/0, f/1, g/1}Property / Task
Prove or disprove commutation.Answer / Result
Yes.Proof (by ACP @ CoCo 2023)
1 Development Closed
Commutation is proven since the TRSs are development closed.
The joins can be performed using
5
step(s).