Certification Problem
Input (COPS 1099)
We consider two TRSs R and S where R contains the rules
|
f(c) |
→ |
g(c) |
(1) |
|
g(c) |
→ |
f(c) |
(2) |
| c |
→ |
d |
(3) |
and S contains the following rules:
| b |
→ |
a |
(4) |
| b |
→ |
c |
(5) |
| c |
→ |
h(b) |
(6) |
| c |
→ |
d |
(3) |
| a |
→ |
h(a) |
(7) |
| d |
→ |
h(d) |
(8) |
The underlying signature is as follows:
{a/0, b/0, c/0, d/0, f/1, g/1, h/1}Property / Task
Prove or disprove commutation.Answer / Result
No.Proof (by ACP @ CoCo 2023)
1 Non-Joinable Fork
The systems are not commuting due to the following forking derivations.
| t0
|
= |
f(c) |
|
→S
|
f(h(b)) |
|
= |
s1
|
and
There is no possibility to join
s1→R*·←S*
t1
for the following reason:
- When applying the cap-function on both terms (where variables may be treated like constants)
then the resulting terms do not unify.