Certification Problem
Input (COPS 1091)
We consider two TRSs R and S where R contains the rules
|
f(a) |
→ |
c |
(1) |
| b |
→ |
c |
(2) |
| b |
→ |
a |
(3) |
and S contains the following rules:
| c |
→ |
f(c) |
(4) |
| c |
→ |
c |
(5) |
|
f(a) |
→ |
b |
(6) |
| a |
→ |
c |
(7) |
The underlying signature is as follows:
{a/0, b/0, c/0, f/1}Property / Task
Prove or disprove commutation.Answer / Result
Yes.Proof (by ACP @ CoCo 2023)
1 Swap TRSs
The role of TRSs R and S are changed.
1.1 Development Closed
Commutation is proven since the TRSs are development closed.
The joins can be performed using
5
step(s).